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The equation formed by multiplying each root of `ax^(2) + bx+ c = 0" by "2 " is "x^(2) = 36x + 24 =0`
If `4^(x)-6.2^(x)8=0`, then the values of x arre

A

`1,2`

B

`1,1`

C

`1,0`

D

`2,2`

Text Solution

Verified by Experts

The correct Answer is:
A
Given equation is.
` 4^(x) - 6.2^(x) +8 = 0 `
` rArr (2^(x))^(2) - 6.2 ^(x) + 8 = 0 `
Put ` 2^(x) = y, ` we have
` y^(2) - 6y + 8 = 0 `
` rArr (y-4) (y-2) = 0 `
` rArr y=2, 4 `
` So, 2^(x) = 2 rArr x = 1 `
` 2^(x) = 2^(2) rArr x = 2 `
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