Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 73 of Exercise 10A
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Solve each of the following quadratic equations:
22x - 3.2(x + 2) + 32 = 0
Given: 22x - 3.2(x + 2) + 32 = 0
(2x)2 - 3.2x.22 + 32 = 0 - - - - (1)
Let 2x = y - - - - - - (2)
substituting for y in (1)
y2 - 12y + 32 = 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 = - 12 c = 32
= 1.32 = 32
And either of their sum or difference = b
= - 12
Thus the two terms are - 8 and - 4
Sum = - 8 - 4 = - 12
Product = - 8. - 4 = 32
y2 - 8y - 4y + 32 = 0
y(y - 8) - 4(y - 8) = 0
(y - 8) (y - 4) = 0
(y - 8) = 0 or (y - 4) = 0
y = 8 or y = 4
2x = 8 or 2x = 4
substituting the value of y in (2)
2x = 23 or 2x = 22
x = 2 or x = 3
Hence the roots of equation are 2, 3
