## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 45 of Exercise 10A

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45
##### Solve each of the following quadratic equations:x2 + 6x - (a2 + 2a - 8) = 0

x2 + 6x - (a2 + 2a - 8) = 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 = 1; b = 6 ;c = - (a2 + 2a - 8)

= 1. - (a2 + 2a - 8)

= - (a2 + 2a - 8)

And either of their sum or difference = b

= 6

Thus the two terms are (a + 4) and - (a - 2)

Difference = a + 4 - a + 2

= 6

Product = (a + 4) - (a - 2)

= - (a2 + 2a - 8)

x2 + 6x - (a2 + 2a - 8) = 0

x2 + (a + 4)x - (a - 2)x - (a + 4)(a - 2) = 0

x [x + (a + 4)] - (a - 2) [x + (a + 4)] = 0

[x + (a + 4)] [x - (a - 2)] = 0

[x + (a + 4)] = 0 or [x - (a - 2)] = 0

x = - (a + 4) or x = (a - 2)

Hence the roots of equation are - (a + 4) or (a - 2)

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