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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`
`(x+1)^(2)-1=0` has

A

one real root

B

two real roots

C

two imaginary roots

D

four real roots

Text Solution

Verified by Experts

The correct Answer is:
B
` (x+1)^(2) - 1 = 0 rArr x+ 1 = pm 1`
` rArr x+1 = 1 or x+1 = - 1 `
` rArr x=0 or x=-2`
Thus `x=0, -2` two roots.
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