Question

If $a{x}^2+2bx+c=0$ and $p{x}^2+2qx+r=0$ have one and only one root in common and $a,b,c,p,q,r$ are rational, then $b^2-ac$ and $q^2-pr$ are

• A. both perfect squares
• B. $b^2-ac$ is a perfect square but $q^2-pr$ is not a perfect square
• C. $q^2-pr$ is a perfect square but $b^2-ac$ is not a perfect square
• D. both are not perfect squares

Answer 1

The correct option is A both perfect squares

Roots of $a{x}^2+2bx+c=0$ are $x=\frac{-2b\pm \sqrt{4b^2-4ac}}{2}=-b\pm \sqrt{b^2-ac}$
And roots of $p{x}^2+2qx+r=0$ are $x=\frac{-2q\pm \sqrt{4q^2-4pr}}{2}=-q\pm \sqrt{q^2-pr}$
Now, we know that irrational roots of quadratic occurs in a pair i.e. if roots will be irrational then both of roots of above quadratics may be common
Hence for only and only root to be common we must have both quadratics roots rational.
And for that we must have $b^2-ac$ and $q^2-pr$ a perfect square