Question

20. Solve for 'x' when x2 - 4ax +4a2- 62 = 0
(a) 2a + b.- 2a - b (6) 2a + b, 2a - b
(c) - 2a + b.- 2a -b
(d) none of these​

Answer 1

hey guys here is your answer

x²-4ax-b²+4a²=0

x²-4ax+4a²-b²=0

(x-2a)²=b²

√(x-2a)²=√b²

x-2a=b

x=2a+b

Hence other root

√(x-2a)²=√b²

x-2a= -b

x=2a-b

So the roots are :2a+b and 2a-b

Hope it helps :3

Answer 2

Step-by-step explanation:

it is too hard that I cannot do it