Question

If $a<b<c<d$, then for any real $\lambda$, the quadratic equation $(x-a)(x-c)+\lambda (x-b)(x-d)=0$ has

• A. Real roots
• B. Imaginary roots
• C. Sum of the roots zero
• D. Product of the roots one.

Answer 1

The correct option is A Real roots

$f\left(x\right)=(x-a)(x-c)+\lambda (x-b)(x-d)$

Now, $f\left(a\right)=\lambda (a-b)(a-d)$ and $f\left(c\right)=\lambda (c-b)(c-d)$

So, $f\left(a\right).f\left(c\right)={\lambda }^2(a-b)(a-d)(c-b)(c-d)<0$

$(\because a<b<c<d)$

Hence, $f\left(x\right)=0$ has a root in $(a,c)$.

$\therefore$ Both the roots of $f\left(x\right)=0$ will be real.

Or we can solve the given problem using discriminant.