Let the quadratic equation be (b+c−a)x2+(c+a−b)x+(a+b−c)=0, where a,b,c∈R and a≠c. 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+c−a)x2+(c+a−b)x+(a+b−c)=0
We know that a+b+c=0
So, (−2a)x2+(−2b)x+(−2c)=0 ⇒ax2+bx+c=0
D=b2−4ac=(−c−a)2−4ac=(c−a)2>0(∵a≠c)∴D>0 ⇒D>0 as a≠c
Hence, the roots are real and distinct.
Alternate solution: (b+c−a)x2+(c+a−b)x+(a+b−c)=0
By observation, one root is x=1 b+c−a+c+a−b+a+b−c=a+b+c=0
Let the other root be α,
Now, the product of roots α×1=a+b−cb+c−a⇒α=−2c−2a=ca
As c≠a, so the equation has real and distinct roots.