{"id":12077,"date":"2020-10-07T17:35:42","date_gmt":"2020-10-07T12:05:42","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=12077"},"modified":"2022-03-02T11:33:50","modified_gmt":"2022-03-02T06:03:50","slug":"quadratic-equations-class-10-extra-questions","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/quadratic-equations-class-10-extra-questions\/","title":{"rendered":"Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Solutions Answers"},"content":{"rendered":"

Here you will find Quadratic Equations\u00a0Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths<\/a> are solved by experts and will guide students in the right direction.<\/p>\n

Extra Questions for Class 10 Maths Quadratic Equations with Answers Solutions<\/h2>\n

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with Solutions Answers<\/strong><\/p>\n

Quadratic Equations Class 10 Extra Questions Objective Type<\/h3>\n

Question 1.
\nThe solution of quadratic equation x2<\/sup> – x – 2 = 0 are:
\n(a) 1, -2
\n(b) -1, -2
\n(c) -1, 2
\n(d) 1, 2.
\nAnswer:
\n(c) -1, 2<\/p>\n

Question 2.
\nThe zero of the polynomial x2<\/sup> + 2x – 3 are:
\n(a) 1, -3
\n(b) -1, 3
\n(c) -1, -3
\n(d) 1, 3
\nAnswer:
\n(a) 1, -3<\/p>\n

\"Quadratic<\/p>\n

Question 3.
\nThe degree of the polynomial x3<\/sup> – x + 7 is:
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) none of these.
\nAnswer:
\n(c) 3<\/p>\n

Question 4.
\nThe zero of the polynomial p(x) = x2<\/sup> + 1 are:
\n(a) real
\n(b) not real
\n(c) (a) and (b) both
\n(d) none of these.
\nAnswer:
\n(b) not real<\/p>\n

Question 5.
\n(i) Every quadratic polynomial can have at the most:
\n(a) one zero
\n(b) two zeroes
\n(c) three zeroes
\n(d) none of these.
\nAnswer:
\n(b) two zeroes<\/p>\n

(ii) From the equation 4\u221a5 x2<\/sup> + 7x – 3\u221a5 = 0, the value of x will be:
\n(a) \"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-1\"
\n(b) \"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-2\"
\n(c) \"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-3\"
\n(d) \"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-4\"
\nAnswer:
\n(a) \"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-1<\/p>\n

Question 6.
\nIf one root of quadratic equation x2<\/sup> + kx + 3 = 0 is 1, then the value of k will be:
\n(a) 1
\n(b) -3
\n(c) -4
\n(d) -5.
\nAnswer:
\n(c) -4<\/p>\n

\"Quadratic<\/p>\n

Question 7.
\nIf one root of quadratic equation 2x2<\/sup> + px – 4 = 0 is 2, then the value of p will be:
\n(a) -3
\n(b) -2
\n(c) 2
\n(d) 3.
\nAnswer:
\n(b) -2<\/p>\n

Question 8.
\nIf one root of quadratic equation ax2<\/sup> + bx + c = 0 is 1, then:
\n(a) a = 1
\n(b) b = 1
\n(c) c = 1
\n(d) a + b + c = 0
\nAnswer:
\n(d) a + b + c = 0<\/p>\n

Question 9.
\nIf 2x2<\/sup> + 1 = 33, then the value of x will be:
\n(a) \u00b12
\n(b) \u00b13
\n(c) \u00b14
\n(d) \u00b11.
\nAnswer:
\n(c) \u00b14<\/p>\n

Question 10.
\nOne root of quadratic equation x2<\/sup> + 3x – 10 = 0:
\n(a) -2
\n(b) +2
\n(c) 0
\n(d) 1
\nAnswer:
\n(b) +2<\/p>\n

Question 11.
\nIf = \\(\\frac {x}{6}\\) = \\(\\frac {6}{x}\\), then the value of x will be:
\n(a) 6
\n(b) -6
\n(c) 6
\n(d) none of these.
\nAnswer:
\n(c) 6<\/p>\n

\"Quadratic<\/p>\n

Question 12.
\nDiscriminant of ax2<\/sup> + bx + c = 0) is:
\n(a) \"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-39\"
\n(b) \"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-40\"
\n(c) \"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-41\"
\n(d) none of these.
\nAnswer:
\n(d) none of these.<\/p>\n

Question 13.
\nA quadratic equation ax2<\/sup> + bx + c = 0, a \u2260 0 has equal roots if b2<\/sup> – 4ac is:
\n(a) equal to 0
\n(b) \u2265 0
\n(c) \u2264 0
\n(d) none of these.
\nAnswer:
\n(a) equal to 0<\/p>\n

Question 14.
\nThe roots of ax2<\/sup> + bx + c = 0, a 50 are real and unequal, if b2<\/sup> – 4ac is:
\n(a) equal to 0
\n(b) \u2265 0
\n(c) \u2264 0
\n(d) none of these.
\nAnswer:
\n(b) \u2265 0<\/p>\n

\"Quadratic<\/p>\n

Question 15.
\nThe roots of the equation 2x2<\/sup> – 8x + c = 0 are eaual, the value of c is:
\n(a) 2
\n(b) 4
\n(c) 6
\n(d) 8
\nAnswer:
\n(d) 8<\/p>\n

Question 16.
\nIf the discriminant (D) of a quadratic equation ax2<\/sup> + bx + c = 0, a \u2260 0 is greater than zero, the roots are:
\n(a) real and unequal
\n(b) real and equal
\n(c) not real
\n(d) none of these.
\nAnswer:
\n(a) real and unequal<\/p>\n

Question 17.
\nAn expression in \u03b1 and \u03b2 is called symmetrical expression if by interchanging \u03b1 and \u03b2, the expression is:
\n(a) changed
\n(b) not changed
\n(c) may be (a) and (b)
\n(d) none of these.
\nAnswer:
\n(b) not changed<\/p>\n

Question 18.
\nIf x2<\/sup> + 5bx + 16 = 0 has no real roots, then:
\n(a) b > \\(\\frac {8}{5}\\)
\n(b) b < \\(\\frac {-8}{5}\\)
\n(c) \\(\\frac {-8}{5}\\) < b < \\(\\frac {8}{5}\\)
\n(d) none of these.
\nAnswer:
\n(c) \\(\\frac {-8}{5}\\) < b < \\(\\frac {8}{5}\\)<\/p>\n

\"Quadratic<\/p>\n

Question 19.
\nIf the roots of 5x2<\/sup> – px + 1 = 0 are real and distinct, then:
\n(a) p > \u221a15
\n(b) p > – 2\u221a5
\n(c) -2\u221a5 < p < 2\u221a5
\n(d) p > 2\u221a5 or p < – 2\u221a5.
\nAnswer:
\n(d) p > 2\u221a5 or p < – 2\u221a5<\/p>\n

Question 20.
\nThe sum of the roots of the quadratic equation 3x2<\/sup> + 4x = 0 is:
\n(a) 0
\n(b) –\\(\\frac {3}{4}\\)
\n(c) –\\(\\frac {4}{3}\\)
\n(d) \\(\\frac {4}{3}\\)
\nAnswer:
\n(c) –\\(\\frac {4}{3}\\)<\/p>\n

Question 21.
\nThe product of the roots of quadratic equation 3x2<\/sup> – 4x = 0) is:
\n(a) 0
\n(b) –\\(\\frac {3}{4}\\)
\n(c) –\\(\\frac {4}{3}\\)
\n(d) –\\(\\frac {4}{3}\\)
\nAnswer:
\n(a) 0<\/p>\n

Question 22.
\nIf one root of x2<\/sup> + kx + 3 = 0 is 1, then the value of k will be:
\n(a) -4
\n(b) -3
\n(c) 1
\n(d) 5.
\nAnswer:
\n(b) -3<\/p>\n

\"Quadratic<\/p>\n

Question 23.
\nIf the sum of the roots of the equation 3x2<\/sup> + (2k + 1)x – (k + 5) = 0 be equal to their product, the value of k is:
\n(a) 5
\n(b) 4
\n(c) 3
\n(d) 2
\nAnswer:
\n(b) 4<\/p>\n

Question 24.
\nThe sum of the roots of quadratic equation 5 – 7x + 3x2<\/sup> = 0 is:
\n(a) +\\(\\frac {7}{5}\\)
\n(b) –\\(\\frac {7}{5}\\)
\n(c) –\\(\\frac {7}{3}\\)
\n(d) +\\(\\frac {7}{3}\\)
\nAnswer:
\n(d) +\\(\\frac {7}{3}\\)<\/p>\n

Question 25.
\nIf one root of quadratic equation x3<\/sup> – 3x + 2 = 0 is 2, then the second root is:
\n(a) 3
\n(b) -1
\n(c) 1
\n(d) 2
\nAnswer:
\n(c) 1<\/p>\n

Quadratic Equations Class 10 Extra Questions Very Short Answer Type<\/h3>\n

Question 1.
\nIf one root quadratic equation x2<\/sup> + 2x – p = 0 is -2, then find the value of p.
\nSolution.
\nGiven that – 2 is a root of given quadratic equation x2<\/sup> + 2x – p = 0
\n\u2234 (-2)2<\/sup> + 2 (-2) – p = 0
\n\u21d2 4 – 4 – p = 0
\n\u2234 p = 0
\nHence, p = 0<\/p>\n

\"Quadratic<\/p>\n

Question 2.
\nIf x2<\/sup> – 5 = 0, then find the value of x:
\nSolution.
\nx2<\/sup> – \\(\\frac {1}{9}\\) = 0
\nx = \u00b1\\(\\frac {1}{3}\\)<\/p>\n

Question 3.
\nIf x = 12 and y = 5, then find the value of 9x2<\/sup> + 49y2<\/sup> – 42xy.
\nSolution.
\nWe have:
\n9x2<\/sup> + 49y2<\/sup> – 42xy
\n= (3x)2<\/sup> + (7y)2<\/sup> – 2.3x.7y
\n= (3x – 7y)2<\/sup>
\n= (3.12 – 7.5)2<\/sup>
\n= (36 – 35)2<\/sup> = 1<\/p>\n

Question 4.
\nSolve the following quadratic equation for x:
\n4x2<\/sup> + 46x – (a2<\/sup> – 12) = 0
\nSolution.
\n4x2<\/sup> + 4bx – a2<\/sup> + b2<\/sup> = 0
\n\u21d2 (2x)2<\/sup> + 2.2xb + (b)2<\/sup>-(a)2<\/sup> = 0
\n\u21d2 (2x + b)2<\/sup>-(a)2<\/sup> = 0
\n\u21d2 (2x + b + a) (2x + b – a) = 0
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-5\"<\/p>\n

\"Quadratic<\/p>\n

Question 5.
\nFind the roots of the quadratic equation:
\n3x2<\/sup> – 2 \u221a6 x + 2 = 0
\nSolution.
\nGiven quadratic equation
\n3x2<\/sup> – 2\u221a6 x + 2 = 0
\n= (\u221a3x)2<\/sup> – 2\u221a3x \u00d7 \u221a2 +(\u221a2)2<\/sup> = 0
\n(\u221a3x – \u221a2)2<\/sup> = 0
\n\u2234 roots are \\(\\frac{\\sqrt{2}}{\\sqrt{3}}\\) or \\(\\frac{\\sqrt{2}}{\\sqrt{3}}\\)<\/p>\n

Question 6.
\nFind the equation whose roots are b – 2a and b + 2a.
\nSolution.
\n\u03b1 = b – 2a
\n\u03b2 = b + 2a
\nSum of roots = \u03b1 + \u03b2 = 6-2a +b + 2a = 2b.
\nProduct of roots = \u03b1\u00df = (b – 2a)(b + 2a)
\n= b2<\/sup> – (2a)2<\/sup>
\n= b2<\/sup> – 4a2<\/sup>
\nThe required quadratic equation is
\n\u2234 x2<\/sup> – (sum of roots) x + product of roots = 0
\n\u2234 x2<\/sup> – (2b)x + (b2<\/sup> – 4a2<\/sup>) = 0
\n\u2234 x2<\/sup> – 2bx + b2<\/sup> – 4a2<\/sup> = 0<\/p>\n

\"Quadratic<\/p>\n

Question 7.
\nIf \u03b1 and \u03b2 are the roots of the equation x2<\/sup> – x – 2 = 0, find that quadratic equation whose roots are (2\u03b1 + 1) and (2\u03b2 + 1).
\nSolution.
\nWe have
\nx2<\/sup> – x – 2 = 0
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-6\"
\nThe required equation is is
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-7\"<\/p>\n

Question 8.
\nIf – 5 is a root of quadratic equation 2x2<\/sup> + px – 15 = 0 and the quadratic equation p(x2<\/sup> + x) + k = 0 has equal roots. Find the value of k.
\nSolution.
\nGiven – 5 is a root of quadratic equation 2x2<\/sup> + px – 15 = 0
\n\u2234 2(-5)2<\/sup> + p(-5) – 15 = 0
\n50 – 5p – 15 = 0
\n5p = 35 \u21d2 p = 7
\nPutting the value of p = 7 in given (ii) quadratic equation.
\n7x2<\/sup> + 7x + k = 0
\nas the roots of this equation are equal
\n\u2234 b2<\/sup> = 4ac
\n\u21d2 (7)2<\/sup> = 4 \u00d7 7 \u00d7 k
\n\u2234 k = \\(\\frac {7}{4}\\)<\/p>\n

\"Quadratic<\/p>\n

Question 9.
\nIf aand Bare the roots of the equation ax2<\/sup> + bx + b = 0, then prove that:
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-8\"
\nSolution.
\nSince, a and \u00df are the roots of the equation
\nax2<\/sup> + bx + b = 0
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-9\"
\nProved.<\/p>\n

Question 10.
\nThe length and breadth of a room are 15 m and 12 m. There is a verandah surrounding the room and area of this is 90m2<\/sup>. Find the width of verandah.
\nSolution.
\nSuppose the width of Verandah is x metre.
\nLength of verandah = (2x + 15) m.
\nBreadth of verandah = (2x + 12) m.
\nArea of verandah = (2x + 15) (2x + 12) – 15 \u00d7 12 = 90
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-10\"
\n\u2234 4x2<\/sup> + 24x + 30x + 180 – 180 – 90 = 0
\n\u2234 4x2<\/sup> + 54x – 90 = 0
\n\u2234 2x2<\/sup> + 27x – 45 = 0
\n\u2234 2x2<\/sup> + 30x – 3x – 45 = 0
\n\u2234 2x(x + 15) – 3(x + 15) = 0
\n(x + 15) (2x – 3) = 0
\nx + 15 = 0
\nor 2x – 3 = 0
\n2x = 3
\nx = 3\/2
\nThe width of verandah = 3\/2 m.<\/p>\n

\"Quadratic<\/p>\n

Question 11.
\nThe square of a positive integer is greater than 11 times the integer by 26. Find the positive integer.
\nSolution.
\nLet the positive integer be x then according to the given condition,
\nx2<\/sup> = 11x + 26
\n\u21d2 x2<\/sup> – 11x – 26 = 0
\n\u21d2 x2<\/sup> – 13x + 2x – 26 = 0
\n\u21d2 x(x – 13) + 2(x – 13) = 0
\n\u21d2 (x – 13)(x + 2) = 0
\n\u2234 x = -2 or x = 13
\nNeglecting the negative value of x, we have x = 13.
\nHence, the integer is 13.<\/p>\n

Question 12.
\nA dealer sells an article for \u20b9 24 and gains as much per cent as the cost price of the article. Find the cost price of the article.
\nSolution.
\nLet cost price of article be \u20b9 x. Then, gain = x%.
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-11\"
\n\u21d2 x2<\/sup> + 100x – 2400 = 0
\n\u21d2 (x + 120)(x – 20) = 0
\n\u21d2 x = -120 or x = 20
\n\u21d2\u00a0 x = 20
\n[\u2234 Cost Price can never be negative]
\nHence, the cost price of the article is \u20b9 20.<\/p>\n

Quadratic Equations Class 10 Extra Questions Short Answer Type<\/h3>\n

Question 1.
\nThe area of a right angle triangle is 30cm2<\/sup>. Find the lengths of the base if its height is 7 cm more then base.
\nSolution.
\nLet the length of base of triangle = x cm
\n\u2234 length of its height = (x + 7) cm
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-12\"
\nArea of right angle \u2206 = \\(\\frac {1}{2}\\) \u00d7 base \u00d7 height
\n\u21d2 30 = \\(\\frac {1}{2}\\)x.(x + 7)
\n\u21d2 x2<\/sup> + 7x – 60 = 0
\n\u21d2 x2<\/sup> + (12 – 5)x – 60 = 0
\n\u21d2 x2<\/sup> + 12x – 5x – 60 = 0
\n\u21d2 x(x +12) – 5(x + 12) = 0
\n\u21d2 (x + 12) (x – 5) = 0
\n\u2234 x = 5 or – 12
\nas length is never negative
\n\u2234 length of base = 5 cm.<\/p>\n

\"Quadratic<\/p>\n

Question 2.
\nThe altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
\nSolution.
\nGiven that, hypotenuse of right triangle = 13 cm
\nLet the base of the right triangle = x
\nAccording to the question,
\nAltitude of the triangle = 7 cm
\nLess than its base = (x – 7) cm
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-13\"
\nBy Pythagoras theorem, we have
\nIn \u2206ABC AC2<\/sup> = BC2<\/sup> + Ab2<\/sup>
\n\u21d2 (13)2<\/sup> = x2<\/sup> + (x – 7)2<\/sup>
\n\u21d2 169 = x2<\/sup> + x2<\/sup> – 14x + 49
\n[\u2235 (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n\u21d2 2x2<\/sup> – 14x – 120 = 0
\n\u21d2 x2<\/sup> – 7x – 60 = 0
\nx2<\/sup> – (12x – 5x) – 60 = 0
\n\u21d2 x2<\/sup> – 12x + 5x – 60 = 0
\n\u21d2 x(x – 12) + 5(x – 12) = 0
\n\u21d2 (x – 12) (x + 5) = 0
\nNow, x – 12 = 0 x = 12
\nand x + 5 = 0 = x = -5
\nSince, altitude of the triangle cannot be negative, hence x \u2260 -5
\nHence, base of the triangle = 12 cm
\nand altitude of the triangle = 12 – 7 = 5 cm.<\/p>\n

\"Quadratic<\/p>\n

Question 3.
\nSolve the equation
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-14\"
\nSolution.
\nGiven,
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-15\"
\n\u21d2 (3x + 4)(x + 4) = 4(x + 1)(x + 2)
\n\u21d2 3x2<\/sup> + 16x + 16 = 4(x2<\/sup> + 3x + 2)
\n\u21d2 3x2<\/sup> + 16x + 16 = 4x2<\/sup> + 12x + 8
\n\u21d2 x2<\/sup> – 4x – 8 = 0
\n\u21d2 Here a =1, b = -4, and c = -8.
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-16\"<\/p>\n

\"Quadratic<\/p>\n

Question 4.
\nA motor boat whose speed in still water is 24 km\/h. takes 1 hour more to go 32 km up stream than to return down stream to the same spot. Find the speed of stream.
\nSolution.
\nLet the speed of stream be x km\/h.
\nGiven that the speed of boat in still water is 24 km\/h
\n\u2234 Speed of boat in down stream = (24 + x) km\/h
\nand speed of boat in up stream = (24 – x) km\/h
\nTime taken by boat to travel 32 km in down
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-17\"
\n\u21d2 32 \u00d7 2x = (24 – x) (24 + x)
\n\u21d2 64x = 576 – x2<\/sup>
\n\u21d2 x2<\/sup> + 64x – 576 = 0
\n\u21d2 x2<\/sup> + (72 – 8)x – 576 = 0
\n\u21d2 x2<\/sup> + 72x – 8x – 576 = 0
\n\u21d2 x(x + 72) – 8(x + 72) = 0 (x – 8)(x + 72) = 0
\n\u2234 x = 8 or -72
\nas speed is never negative
\n\u2234 x = 8
\ni.e., speed of stream = 8 km\/h.<\/p>\n

\"Quadratic<\/p>\n

Question 5.
\nSolve the quadratic equation
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-18\"
\nby factorization method
\nSolution.
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-19\"
\n\u21d2 x(a + b + x) = – ab
\n\u21d2 ax + bx + x2<\/sup> + ab = 0
\n\u21d2 x2<\/sup> + ax + bx + ab = 0
\n\u21d2 x(x + a) + b (x + a) = 0
\n\u21d2 (x + a) (x + b) = 0
\n\u2234 x = -a or -6.<\/p>\n

\"Quadratic<\/p>\n

Question 6.
\nIf the equations x2<\/sup> + kx + 64 = 0 and x2<\/sup> – 8x + k = 0 have real roots. Find the positive value of k.
\nSolution.
\nGiven quadratic equation are
\nx2<\/sup> + kx + 64 = 0. …(i)
\nand x2<\/sup> – 8x + k = 0…(ii)
\nIf the equations have real roots, then D \u2265 0 Discriminant of (i) k2<\/sup> – 256 and equation<\/p>\n

(ii) 64 – 4k
\n= k2<\/sup> – 256 \u2265 0 and 64 – 4k \u2265 0
\n\u21d2 k2<\/sup> \u2265 256 and 64 \u2265 4k
\n\u21d2 k \u2265 16 and 16 \u2265 k.
\nHence, k = 16.<\/p>\n

\"Quadratic<\/p>\n

Question 7.
\nIf the speed of a train is increased by 5 km\/H then train takes 1 hour less time to cover a distance of 360 km. Find the speed of the train.
\nSolution.
\nLet the speed of train be x km\/ Hr.
\n\u2234 Time taken by the train to cover 360
\nkm = \\(\\frac {360}{x}\\) Hours
\nNew speed of the train = (x + 5) km\/Hr.
\n\u2234 New time taken to cover 360 km
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-20\"
\n\u21d2 360 \u00d7 5 = x2<\/sup> + 5x
\n\u21d2 x2<\/sup> + 5x – 1800 = 0
\n\u21d2 x2<\/sup> + 45x – 40x – 1800 = 0
\n\u21d2 x(x + 45) – 40(x + 45) = 0
\n\u21d2 (x + 45) (x – 40) = 0
\n\u2234 x = 40 or -45
\nas speed is never negative
\n\u2234 speed of train = 40 km\/hour.<\/p>\n

\"Quadratic<\/p>\n

Question 8.
\nSolve the following equation
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-21\"
\nSolution.
\nLet = \\(\\frac{x-2}{x+2}\\) = y
\n\u2234 Equation reduced to
\ny + \\(\\frac {3}{y}\\) – 4 = 0
\n\u21d2 y2<\/sup> + 3 – 4y = 0
\n\u21d2 y2<\/sup> – 4y + 3 = 0
\n\u21d2 y2<\/sup> – 3y – y + 3 = 0
\n\u21d2 y(y – 3)-1 (y – 3) = 0
\n\u21d2 (y – 3)(y – 1) = 0
\n\u2234 y = 3 or 1.
\nWhere
\ny = 3 = \\(\\frac{x-2}{x+2}\\)
\n3x + 6 = x – 2
\n\u21d2 2x = -8
\n\u2234 x = -4
\nWhere
\ny = 1 = \\(\\frac{x-2}{x+2}\\)
\nx + 2 x + 2 = x – 2 which is not true
\n\u2234 x = -4.<\/p>\n

\"Quadratic<\/p>\n

Question 9.
\nThe sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is a. Find his present age.
\nSolution.
\nLet the present age of Rehman – x year
\n\u2234 Rehman’s age, 3 yr ago = (x – 3) yr
\nRehman’s age, after 5 yr = (x + 5) yr
\nAccording to the question,
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-22\"
\n\u2234 6x + 6 = x2<\/sup> + 2x – 15
\n\u21d2 x2<\/sup> + 2x – 6x – 15 – 6 = 0
\nx2<\/sup> – 4x \u2212 21 = 0
\nx2<\/sup> – (7x – 3x) – 21 = 0
\n\u21d2 x2<\/sup> – 7x + 3x – 21 = 0
\n(By factorization method)
\n\u21d2 x(x – 7) + 3(x – 7) = 0
\n\u21d2 (x – 7)(x + 3) = 0
\n\u2234 x – 7 = 0 \u21d2 x = 7 or x + 3 = 0 \u21d2 x = -3
\nBut x +-3 age cannot be negative.
\nSo, present age of Rehman = 7 yr.<\/p>\n

\"Quadratic<\/p>\n

Question 10.
\nThe difference of two numbers is 2 and the sum of their square is 34. Find the number.
\nSolution.
\nLet Numbers be x and y, such x > y
\n\u2234 x – y = 2 …..(i)
\nand x2<\/sup> + y2<\/sup> = 34 …(ii)
\nfrom (i) x = 2 + y
\n\u21d2 (2 + y)2<\/sup> + y2<\/sup> = 34
\n\u21d2 2y2<\/sup> + 4y – 30 = 0
\n\u21d2 y2<\/sup> + 2y – 15 = 0
\n\u21d2 (y + 5) (y – 3) = 0
\n\u2234 y = 3 or – 5
\nwhere y = 3, x = 2 + 3 = 5
\n\u2234 Numbers are 5 and 3.<\/p>\n

\"Quadratic<\/p>\n

Question 11.
\nThe speed of boat in still water is 15km\/H. The boat completes 30 km distance in going downstream and 30 km up stream in total 4 Hr 30 minute. Find the speed of stream.
\nSolution.
\nLet the speed of stream = x km\/ Hr.
\n\u2234 speed of boat in down stream = (15 + x) km\/H
\nand speed of boat in up stream = (15 – x) km\/H
\nTime taken 30 km in down stream = \\(\\frac{30}{(15+x)}\\) Hrs
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-23\"
\n\u21d2 2 \u00d7 30 \u00d7 30 = 9(225 – x2<\/sup>)
\n\u21d2 200 = 225 – x2<\/sup>
\n\u21d2 x2<\/sup> = 225 – 200
\n= 25
\nx = \u221a25 = \u00b1 5
\nas speed is never negative:. speed to stream = 5 km\/Hr.<\/p>\n

\"Quadratic<\/p>\n

Question 12.
\nDetermine whether the quadratic equation 9x2<\/sup> + 7x – 2 = 0 has real roots. If so, find roots of quadratic equation.
\nSolution.
\nGiven, quadratic equation
\n9x2<\/sup> + 7x – 2 = 0
\nHere a = 9, b = 7 and c= -2
\nD = b2<\/sup> – 4ac .
\n= (7)2<\/sup> – 4 \u00d7 9 \u00d7 -2
\n= 49 + 72 = 121
\n\u2234 roots of quadratic equation are real.
\n\u2234 roots are
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-24\"<\/p>\n

\"Quadratic<\/p>\n

Question 13.
\nBy solving the equation
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-25\"
\nget a quadratic equation. Find the nature of roots. Using formula, solve the quadratic equation.
\nSolution.
\nGiven,
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-26\"<\/p>\n

Nature of root, D = b2<\/sup> – 4ac = (4)2<\/sup> – 4 \u00d7 1 \u00d7 0 = 16 > 0
\n\u2234 root of the equation are real.
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-27\"<\/p>\n

\"Quadratic<\/p>\n

Question 14.
\nSome vessels are manufactured in a day in a small industry. On a particular day it is observed that the cost of manufacturing each item was 3 more than twice the number of vessels manufactured on that day. If the total manufacturing cost on that day was 90, then find the number of manufactured vessels and manufacturing cost of each item.
\nSolution.
\nLet Number of vessel manufacture in a day = x
\n\u2234 Cost of manufacturing = (2x + 3).
\nAccording to Question,
\nx(2x +3) = 90
\n\u21d2 2x2<\/sup> + 3x – 90 = 0
\n\u21d2 2x2<\/sup> + 15x – 12x – 90 = 0
\n\u21d2 x( 2x + 15)-6 (2x + 15) = 0
\n\u21d2 (x – 6) (2x + 15) = 0
\n\u2234 x = 6 or – \\(\\frac {15}{2}\\)
\nas number of vessels are not negative
\n\u2234 x = 6
\nAlso the manufacturing cost = 2 \u00d7 6 + 3 = \u20b9 15.<\/p>\n

\"Quadratic<\/p>\n

Question 15.
\nIs the following situation possible ? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
\nSolution.
\nLet age of one of the two friends = x yr
\nThen, age of other friend = (20 – x) yr
\n(\u2235 The sum of the ages of two friends is 20 yr)
\n4 yr ago age of one of the two friends = (x – 4) yr
\n4 yr ago age of the other friend
\n= (20 – x – 4) yr = (16 – x) yr
\nAccording to the question,
\n(x – 4)(16 – x) = 48
\n\u21d2 16x – x2<\/sup> – 64 + 4x = 48
\n\u21d2 x2<\/sup> – 20x + 112=0
\nOn comparing the above equation with ax2<\/sup> + bx + c = 0, we get
\na = 1, b = -20 and c = 112
\n\u2234 Discriminant, D = b2<\/sup> – 4ac
\n= (-20)2<\/sup> – 4 \u00d7 1 \u00d7 112
\n= 400 – 448 = 48 < 0
\nwhich implies that the real roots are not possible because this condition represents imaginary roots. So, the solution does not exist.<\/p>\n

\"Quadratic<\/p>\n

Question 16.
\nIs it possible to design a rectangular park of perimeter 80 m and area 400 m2<\/sup> ? If so, find its length and breadth.
\nSolution.
\nLet the breadth of the park = x metre
\nThen, according to the question, Perimeter of a rectangular park = 80 m
\n= 2(Length + Breadth)=80 m
\n= Length + Breadth = 40 m
\nLength = (40 – x) m
\n\u2234 Area of a rectangular park = Length \u00d7 Breadth = (40 – x)x m2<\/sup>
\nBut according to the question,
\nArea of the rectangular park is 400 m2<\/sup>.
\n\u2234 (40 – x)x = 400
\n\u21d2 x2<\/sup> – 40x + 400 = 0
\n\u21d2 x2<\/sup> – 2x \u00d7 20 + (20)2<\/sup> = 0
\n\u21d2 (x – 20)2<\/sup> = 0
\n[\u2235 a2<\/sup> – 2ab + b2<\/sup> = (a – b)2<\/sup>]
\n\u21d2 x = 20
\nThus, breadth of the park = 20 m
\nand length of the park = (40 – 20) m = 20 m
\nSo, it is possible to design the rectangular park of perimeter 80 m and area 400 m2<\/sup>. But this rectangle will be a square of side 20 m.<\/p>\n

Quadratic Equations Class 10 Extra Questions Long Answer Type<\/h3>\n

Question 1.
\nSolve:
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-28\"
\nx \u2260 0.
\nSolution.
\nThe given equation is:
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-29\"
\nTherefore, the given equation can be written as
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-30\"
\nNow, substituting y = x + \\(\\frac {1}{x}\\), we get
\n4y2<\/sup> + 8y – 45 = 0
\n\u21d2 4y2<\/sup> + 18y – 10y – 45 = 0
\n\u21d2 2y (2y + 9) – 5(2y +9) = 0
\n\u21d2 (2y + 9) (2y – 5) = 0
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-31\"
\nNow, from (ii), we get,
\n2(x2<\/sup> + 1) = 5x
\n\u21d2 2x2<\/sup> – 5x + 2 = 0
\n\u21d2 2x2<\/sup> – 4x – x + 2 = 0
\n\u21d2 2x(x – 2) – 1(x – 2) = 0
\n\u21d2 (x – 2)(2x – 1) = 0
\nx = 2 and x = \\(\\frac {1}{2}\\)
\nHence, the solutions of the given equation are
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-32\"<\/p>\n

\"Quadratic<\/p>\n

Question 2.
\nSolve: 2x4<\/sup> – x3<\/sup> – 11x2<\/sup> + x + 2 = 0.
\nSolution.
\nGiven equation is
\n2x4<\/sup> – x3<\/sup> – 11x2<\/sup> – x + 2 = 0
\nDivide both sides by x2<\/sup>,
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-33\"
\nNow substitute x + \\(\\frac {1}{x}\\) = y
\n\u2234 2y2<\/sup> – y – 15 = 0
\n\u21d2 2y2<\/sup> – 6y + 5y – 15 = 0
\n\u21d2 2y(y – 3) + 5(y – 3) = 0
\n\u21d2 (y – 3)(2y + 5) = 0
\n\u2234 y = 3 or y = –\\(\\frac{5}{2}\\)
\ny = 3 = \\(\\frac{x^{2}+1}{x}\\)
\nWhen
\n\u21d2 x2<\/sup> + 1 = 3x
\n\u21d2 x2<\/sup> – 3x + 1 = 0
\nHere, a = 1, b = -3 and c = 1
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-34\"
\n\u21d2 2x2<\/sup> + 5x + 2 = 0
\n\u21d2 2x2<\/sup> + 4x + x + 2 = 0
\n\u21d2 2x(x + 2) + 1(x + 2) = 0
\n\u21d2 (x + 2)(2x + 1) = 0
\nx = -2, – \\(\\frac{1}{2}\\)
\nHence, the solutions are -2, –\\(\\frac{1}{2}\\), \\(\\frac{3 \\pm \\sqrt{5}}{2}\\).<\/p>\n

\"Quadratic<\/p>\n

Question 3.
\nSolve \\(\\sqrt{3 x+10}\\) + \\(\\sqrt{6-x}\\) = 6.
\nSolution.
\nGiven equation is:
\n\\(\\sqrt{3 x+10}\\) + \\(\\sqrt{6-x}\\) = 6.
\nWe must look for solutions which satisfy
\n3x + 10 \u2265 0 and 6 – x \u2265 0
\nx \u2265 –\\(\\frac {10}{3}\\) and x \u2264 6
\ni.e., –\\(\\frac {10}{3}\\) \u2264 x \u2264 6.
\nNow, we first transform one of the radicals co the R.H.S.
\nWe have,
\n\\(\\sqrt{3 x+10}\\) = 6 – \\(\\sqrt{6-x}\\)
\nSquaring both sides, we get
\n3x + 10 = 36 + 6 – x – 12\\(\\sqrt{6-x}\\)
\nor 4x – 32 = -127\\(\\sqrt{6-x}\\)
\nor 8 – x = 3\\(\\sqrt{6-x}\\)
\nAgain, squaring both sides, we get
\n64 + x2<\/sup> – 16x = 9 (6 – x)
\nor 64 + x2<\/sup> – 16x – 54 + 9x = 0
\nor x2<\/sup> – 7x + 10 = 0
\nor (x \u2212 2)(x – 5) = 0
\n\u2234 x = 2 and x = 5.
\nwhich satisfy \\(\\frac{-10}{3}\\) \u2264 x \u2264 6.
\nHence, the solutions are x = 2 and x = 5.<\/p>\n

\"Quadratic<\/p>\n

Question 4.
\nOne person purchased some cloth for \u20b9 2250. Had the price of cloth been \u20b9 50 per metre higher, then he would have been able to purchase the cloth less by 1.5 metre. Find the measure of the cloth purchased and price per metre.
\nSolution.
\nLet the price per metre of cloth be \u20b9 x
\nAmount spend = \u20b9 2250
\n\u2234 Length of cloth purchased
\n\\(\\frac{2250}{x}\\) = metre.
\nNow, new price of cloth per metre become = \u20b9 (x + 50)
\n\u2234 Length of cloth purchase
\n= \\(\\frac{2250}{x + 50}\\) metre.
\nAccording to the question,
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-35\"
\n\u21d2 2250 \u00d7 50 \u00d7 2 = 3(x2<\/sup> + 50x)
\n\u21d2 x2<\/sup> + 50x – 75000 = 0
\n\u21d2 x2<\/sup> + 300x – 250x – 75000 = 0
\n\u21d2 x(x + 300) – 250(x + 300) = 0
\n\u21d2 (x + 300)(x – 250) = 0
\n\u2234 x = -300 or x = 250
\nNeglecting the -ve sign value i.e.,
\nx = -300
\nHence, cost of cloth per metre = \u20b9 250.
\nand length of cloth \\(\\frac{2250}{250}\\) = 9 metre.<\/p>\n

\"Quadratic<\/p>\n

Question 5.
\nA two-digit number is such that the product of the digits is 12, when 36 is added to the number, the digit interchange their places. Find the number.
\nSolution.
\nSuppose two digit number is 10x + y.
\nWhose y = unit’s place and x = ten’s place digit.
\nAccording two question
\nxy = 12 …..(i)
\nand 10x + y + 36 = 10y + x.
\n\u2234 9x – 9y = -36
\n\u2234 x – y = -4
\nx = y – 4
\nPut x = y – 4 in (i)
\n(y – 4)y = 12
\n\u2234 y2<\/sup> – 4y – 12 = 0
\n\u2234 y2<\/sup> – 6y + 2y – 12 = 0
\n\u2234 y(y – 6) + 2(y – 6) = 0
\n\u2234 (y – 6) (y + 2) = 0
\ny – 6 = 0 or y + 2 = 0
\ny = 6 or y = -2
\n(not acceptable).
\n\u2234 Put y = 6 in (i) we get
\nx (6) = 12
\nx = 2
\nThe two digit number is
\n10x + y = 10(2) + 6 = 26.<\/p>\n

\"Quadratic<\/p>\n

Question 6.
\nThe diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, find the sides of the field.
\nSolution.
\nLet PQRS be the rectangular field. Let the shorter side QR of the rectangle = x m.
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-36\"
\nAccording to the question,
\nDiagonal of the rectangle PR = 60 m more than the shorter side = (x + 60) m
\nSide of the rectangle PQ = 30 m more than the shorter side = (x + 30) m
\n\u2234 By Pythagoras theorem,
\nIn \u2206PQR, PR2<\/sup> = PQ2<\/sup> + QR2<\/sup>
\n(\u2235 In rectangle every adjacent side makes an angle 90\u00b0 to each other)
\n\u21d2 (x + 60)2<\/sup> = (x + 30)2<\/sup> + x2<\/sup>
\n\u21d2 x2<\/sup> + 120x + 3600 = x2<\/sup> + 60x + 900 + x2<\/sup>
\n[\u2235 (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]
\n\u21d2 (2x2<\/sup> – x2<\/sup>) + (60x – 120x) + (900 – 3600) = 0
\nx2<\/sup> – 60x – 2700 = 0
\nx2<\/sup> – (90x – 30x) – 2700 = 0
\nx2<\/sup> – 90x + 30x – 2700 = 0
\n(By factorization method)
\n\u21d2 x(x – 90) + 30(x – 90) = 0
\n\u21d2 (x – 90)(x + 30) = 0
\n\u2234 Either x – 90 = 0 \u21d2 x = 90 or x + 30 = 0
\n\u21d2 x = – 30 which is not possible because side cannot be negative.
\n\u2234 x = 90
\nSo, breadth of the rectangle = 90 m and length of the rectangle = 90 + 30 = 120 m.<\/p>\n

\"Quadratic<\/p>\n

Question 7.
\nThe difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
\nSolution.
\nLet the required numbers be x and y, where x > y
\nDifference of squares of two numbers = 180
\nGiven, x2<\/sup> – y2<\/sup> = 180 …(i)
\nSquare of smaller number = 8 \u00d7 Larger number
\n\u21d2 y2<\/sup> = 8x …..(ii)
\nFrom Eqs. (i) and (ii), we have
\nx2<\/sup> – 8x = 180
\n\u21d2 x2<\/sup> – 8x – 180 = 0
\nx2<\/sup> – (18x – 10x) – 180 = 0
\n\u21d2 x2<\/sup> – 18x + 10x – 180 = 0
\n(By factorization method)
\n\u21d2 x(x – 18) + 10(x – 18) = 0
\n\u21d2 (x – 18)(x + 10) = 0
\n\u2234 x – 18 = 0 or x + 10 = 0
\n\u2234 x = 18 or x = – 10
\nNow, x = 18
\n\u21d2 y2<\/sup> = 8 \u00d7 18 = 144
\n\u21d2 y = \u00b1 12
\n\u21d2 y = 12 or -12 [From EQuestion (ii)]
\nAgain, x = -10
\n\u21d2 y2<\/sup> = [8 (-10)] = -80
\nwhich is not possible i.e., imaginary value.
\nHence, the numbers are (18 and 12) or (18 and -12).<\/p>\n

\"Quadratic<\/p>\n

Question 8.
\nAn express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 kmh-1<\/sup> more than that of the passenger train, find the average speed of the two trains.
\nSolution.
\nLet the average speed of the passenger train = x kmh-1<\/sup>.
\nThen, the average speed of the express train = (x + 11) kmh-1<\/sup>
\nTime taken by passenger train to cover
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-37\"
\n\u21d2 x (x + 11) = 132 \u00d7 11
\n\u21d2 x2<\/sup> + 11x – 1452 = 0
\nx2<\/sup> + (44 – 33)x – 1452 = 0
\n\u21d2 x2<\/sup> + 44x – 33x – 1452 = 0
\n(By fatorization method)
\n= x(x + 44) – 33(x + 44) = 0
\n\u21d2 (x + 44)(x – 33) = 0
\n\u2234 x + 44 = 0
\n\u21d2 x = -44 or x – 33 = 0
\n\u21d2 x = 33
\nSince, x \u2260 -44 because speed can’t be negative.
\nHence, the average speed of the passenger train = 33 km\/h
\nand the average speed of the express train = (33 + 11) km\/h = 44 km\/h.<\/p>\n

\"Quadratic<\/p>\n

Question 9.
\nOut of a certain number of Saras birds, one-fourth of the number are moving about in lotus plants, \\(\\frac {1}{9}\\)th coupled with \\(\\frac {1}{4}\\)th as well as 7 times the square root of the number move on a hill, 56 birds remain in vakula tree. What is the total number of birds ?
\nSolution.
\nLet the total number of birds be x.
\n\u2234 Number of birds moving about in lotus plants = \\(\\frac {x}{4}\\)
\nand the number of birds moving on a hill
\n\u21d2 \\(\\frac {x}{9}\\) + \\(\\frac {x}{4}\\) + 7\u221ax
\nAlso, the number of birds in vakula tree = 56
\n\u2234 According to the given information, we have
\n\"Quadratic-Equations-Class-10-Extra-Questions-Maths-Chapter-4-with-Solutions-Answers-38\"
\n9x + 2x + 126 \u221ax – 18x + 1008 = 0
\n\u21d2 -7x + 126 \u221ax + 1008 = 0
\n\u21d2 x – 18\u221ax – 144 = 0
\nPutting \u221ax = y, we get
\ny2<\/sup> – 18y – 144 = 0
\n\u21d2 y2<\/sup> – 24y + 6y – 144 = 0
\n\u21d2 y(y – 24) + 6(y – 24) = 0
\n\u21d2 (y – 24)(y + 6) = 0
\n\u21d2 y = 24 or y = -6.
\nBut y = -6, since \u221ax = y is positive
\n\u2234 y = 24
\n\u21d2 \u221ax = 24
\n\u2234 x = 576.
\nHence, the total number of birds is 576.<\/p>\n

\"Quadratic<\/p>\n

Question 10.
\nIs it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m2<\/sup> ? If so, find its length and breadth.
\nSolution.
\nLet breadth of a rectangular mango grove = x metre
\n\u2234 Length of a rectangular mango grove = 2x metre(By given condition)
\n\u2234 According to the question,
\nArea of rectangular mango grove = 800 m2<\/sup>
\n\u21d2 Length \u00d7 Breadth = 2x (x) = 800
\n\u21d2 2x2<\/sup> = 800
\n\u21d2 x2<\/sup> = 400
\n\u21d2 x = \u00b1 20
\nBut (\u2235 breadth never be negative)
\n\u2234 Length 2x = 40 m and Breadth = 20 m.<\/p>\n

\"Quadratic<\/p>\n","protected":false},"excerpt":{"rendered":"

Here you will find Quadratic Equations\u00a0Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Quadratic Equations with Answers Solutions Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with …<\/p>\n

Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Solutions Answers<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nQuadratic Equations Class 10 Extra Questions Maths Chapter 4 with Solutions Answers - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/quadratic-equations-class-10-extra-questions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Solutions Answers - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"Here you will find Quadratic Equations\u00a0Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. 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