Question

Assertion :If $a+b+c=0$ and $a,b,c$ are rational, then roots of the equation $(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$ are rational. Reason: For quadratic equation given in Assertion, Discriminant is perfect square.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

For the above given quadratic equation, the discriminant will be
$(a+c-b{)}^2-4[(b+(a-c\left)\right)(b-(a-c\left)\right)]$
$=(a+b+c-2b{)}^2-4[(a+b+c-2c)(a+b+c-2a)]$
$=(2b{)}^2-4[(-2c)(-2a)]$ ...since $a+b+c=0$
$=4b^2-4.4ac$
$=4[b^2-4ac]$
$=4\left[\right(-a-c{)}^2-4ac]$
$=4[a^2+c^2+2ac-4ac]$
$=4\left[\right(a-c{)}^2]$
Hence, discriminant is a perfect square.
Therefore, the roots are rational.