Question

If $a{x}^2+bx+c=0$ and $c{x}^2+bx+a=0$ $(a,b,c\in R)$ have a common non - real root, then

• A. $-2|a|<|b|<2|a|$
• B. $-2|c|<|b|<2|c|$
• C. $a=c$
• D. All of these

Answer 1

The Correct Option is D

Explanation:
Given: Quadratic equations, $a{x}^2+bx+c=0$.....(i)and $c{x}^2+bx+a=0$....(ii)
We have to find the condition for which above two quadratic equations have a common non-real root.
For quadratic equation (i), ${\mathrm{D}}_1={\mathrm{b}}^2-4\mathrm{ac}<0$ [As roots are non-real]
For quadratic equation (ii) ${\mathrm{D}}_2={\mathrm{b}}^2-4\mathrm{ac}<0$
Now we know that, both equations has complex roots and complex roots exist in conjugate pairs.If one root is common, then other roots will also be common due to the nature of conjugate roots.
Applying condition of both roots common for quadratic equations,
we get$\frac{a}{c}=\frac{b}{b}=\frac{c}{a}$$-\frac{a}{c}=1=\frac{c}{a}$$-c=a$.....(iii)
Now, $b^2-4ac<0$$-b^2-4a^2<0$ or $b^2-4c^2<0$ [Using (iii)]$-2|a|<|b|<2|a|$ or $-2|c|<|b|<2|c|$
Hence, the correct option is (D).