Question

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

• A. Non-real roots.
• B. One real root between $a$ and $c$.
• C. One real root between $b$ and $d$.
• D. Irrational roots.

Answer 1

Answer: (B), (C)

One real root between $b$ and $d$.
One real root between $a$ and $c$.

Let $f\left(x\right)=(x-a)(x-c)+\lambda (x-b)(x-d)$
$f\left(a\right)=\lambda (a-b)(a-d)$
$f\left(c\right)=\lambda (c-b)(c-d)$
$\therefore f\left(a\right)f\left(c\right)={\lambda }^2(a-b)(a-d)(c-b)(c-d)<0$ $[\because a<b<c<d]$
Hence, $f\left(x\right)=0$ has one real root between $a$ and $c$.
Again, $f\left(b\right)f\left(d\right)=(b-a)(b-c)(d-a)(d-c)<0$.
Hence, $f\left(x\right)=0$ has one real root between $b$ and $d$.