RS Aggarwal - Mathematics

Book: RS Aggarwal - Mathematics

Chapter: 10. Quadratic Equations

Subject: Maths - Class 10th

Q. No. 67 of Exercise 10A

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67

Solve each of the following quadratic equations:

Given:

taking LCM; using (a + b)2 = a2 + b2 + 2ab





- 24x2 - 64x - 1 = 3(8x2 - 2x - 3) cross multiplying


- 24x2 - 64x - 1 = 24x2 - 6x - 9


48 x2 + 58x - 8 = 0 taking 2 common


24 x2 + 29x - 4 = 0


Using the splitting middle term - the middle term of the general equation is divided in two such values that:


Product = a.c


For the given equation a = 24 b = 29 c = - 4


= 24. - 4 = - 96


And either of their sum or difference = b


= 29


Thus the two terms are 32 and - 3


Difference = 32 - 3 = 29


Product = 32. - 3 = - 96


24 x2 + 29x - 4 = 0


24 x2 + 32x - 3x - 4 = 0


8x(3x + 4) - 1(3x + 4) = 0


(3x + 4)(8x - 1) = 0


(3x + 4) = 0 or (8x - 1) = 0



Hence the roots of equation are


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