Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 71 of Exercise 10A
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Solve each of the following quadratic equations:
3(x + 2) + 3 - x = 10
Given: 3(x + 2) + 3 - x = 10
- - - - - - - - (1)
Let 3x = y - - - - - - - - - - (2)
substituting for y in (1)
9y2 - 10y + 1 = 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 = 9 b = - 10 c = 1
= 9.1 = 9
And either of their sum or difference = b
= - 10
Thus the two terms are - 9 and - 1
Sum = - 9 - 1 = - 10
Product = - 9. - 1 = 9
9y2 - 9y - 1y + 1 = 0
9y(y - 1) - 1(y - 1) = 0
(y - 1) (9y - 1) = 0
(y - 1) = 0 or (9y - 1) = 0
y = 1 or y = 1/9
3x = 1 or 3x = 1/9
On putting value of y in equation (2)
3x = 30 or 3x = 3 - 2
x = 0 or x = - 2
Hence the roots of equation are 0, - 2
