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Question

Let the quadratic equation be (b+ca)x2+(c+ab)x+(a+bc)=0, where a,b,cR and ac. If a+b+c=0, then the roots are

A
not real
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B
real and equal
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C
real and distinct
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D
real, equal in magnitude but opposite in sign
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Solution

The correct option is C real and distinct
Given equation is
(b+ca)x2+(c+ab)x+(a+bc)=0
We know that
a+b+c=0
So,
(2a)x2+(2b)x+(2c)=0
ax2+bx+c=0

D=b24ac =(ca)24ac =(ca)2>0 (ac)D>0
D>0 as ac

Hence, the roots are real and distinct.


Alternate solution:
(b+ca)x2+(c+ab)x+(a+bc)=0
By observation, one root is
x=1
b+ca+c+ab+a+bc=a+b+c=0
Let the other root be α,
Now, the product of roots
α×1=a+bcb+caα=2c2a=ca

As ca, so the equation has real and distinct roots.

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