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If the quadratic equations, a x^2+2c x+b...

If the quadratic equations, `a x^2+2c x+b=0a n da x^2+2b x+c=0(b!=c)` have a common root, then `a+4b+4c` is equal to: a. -2 b. -2 c. 0 d. 1

A

`-2`

B

`-1`

C

`0`

D

`1`

Text Solution

Verified by Experts

The correct Answer is:
C
We have `(a.2b-2c.a)(2c.c-b.2b)=(ba-ca)^(2)`
`implies2a(b-c).2(c^(2)-b^(2))=a^(2)(b-c)^(2)`
`implies4a(c-b)(c+b)=a^(2)(b-c)[:'b!=c]`
`implies4a(c+b)=-a^(2)`
`impliesa+4b+4c=0`
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