Question

The quadractic equation $ab{x}^2+acx+b(bx+c)=0$ has non-zero, equal and rational roots.The value of a and c respectively cannot be equal to $(ab\ne 0)$

• A. $4$ & $49$
• B. $4$ &$16$
• C. $4$ & $64$
• D. $8$ &$49$

Answer 1

The correct option is D $8$ &$49$

Consider the quadratic equation

$ab{x}^2+acx+b(bx+c)=0$

$ab{x}^2+(ac+b^2)x+c=0$

Given:Roots are non-zero equal&rational.

$\Rightarrow$ discriminant $=0$

$△=0\Rightarrow B^2-4AC=0$

$\Rightarrow {(ac+b^2)}^2-4\left(ab\right)\left(bc\right)=0$

$\Rightarrow a^2{c}^2+b^4+2a{b}^{2}c-4a{b}^{2}c=0$

$\Rightarrow b^4+a^2{c}^2-2a{b}^{2}c=0$

$\Rightarrow {\left(b^2\right)}^2+{\left(ac\right)}^2-2{\left(b\right)}^2\left(ac\right)=0$

${(b^2-ac)}^2=0$

So $ac=b^2$

We get the condition is $ac=b^2$

Option $\left(a\right)4\times 49={(2\times 7)}^2\Rightarrow b^2={14}^2$ is rational

Option $\left(b\right)4\times 16={(2\times 4)}^2\Rightarrow b^2=8^2$ is rational

Option $\left(c\right)4\times 64={(2\times 8)}^2\Rightarrow b^2={16}^2$

Option $\left(d\right)8\times 49={(2\sqrt{2}\times 7)}^2\Rightarrow b^2={\left(14\sqrt{2}\right)}^2$

$\Rightarrow b=14\sqrt{2}$ is irrational

Hence, $a$ and $c$ respectively cannot be equal to $8$ and $49$