Question

The roots of $(x-a)(x-c)+k(x-b)(x-d)=0$ are real and distinct for all real $k$ if

• A. $a<b<c<d$
• B. $a>b<c<d$
• C. $a<b>c<d$
• D. $a<b=c<d$

Answer 1

The correct option is A $a<b<c<d$

The roots of $(x-a)(x-c)+k(x-b)(x-d)=0$ are real and distinct for all real $k$

$f\left(x\right)=(1+k)x^2-(a+c+k(b+d\left)\right)x+ac+kbd$

As $D>0$ for real roots.

So ,

$\Rightarrow$${(a+c+k(b+d\left)\right)}^2-4(1+k)(ac+kbd)>0$

$\Rightarrow$$(b+d{)}^2{k}^2+2(a+c\left)\right(b+d)k+(a+c{)}^2-4(bd{k}^2+(ac+bd)k+ac)>0$

$\Rightarrow$ $(b-d{)}^2{k}^2+(2(a+c)(b+d)-4(ac+bd))k+(a-c{)}^2>0$

This true for all $k$

$D<0$

$\left(2\right(a+c\left)\right(b+d)-4(ac+bd){)}^2-4((a-c)(b-d){)}^2<0$

$\left(\right(a+c\left)\right(b+d)-2(ac+bd){)}^2-((a-c)(b-d){)}^2<0$

Applying $a^2-b^2=(a-b)(a+b)$ we get,

$\Rightarrow$ $\left(\right(a+c\left)\right(b+d)-2(ac+bd)+(a-c\left)\right(b-d\left)\right)\left(\right(a+c\left)\right(b+d)-2(ac+bd)-(a-c\left)\right(b-d\left)\right)<0$

$\Rightarrow$$(ab+ad+bc+cd-2ac-2bd+ab-ad-bc+cd)(ab+ad+bc+cd-2ac-2bd-ab+ad+bc-cd)<0$

$\Rightarrow$$(ab+cd-ac-bd)(ad+bc-ac-bd)<0$

$\Rightarrow$$(a-d)(b-c)(a-b)(d-c)<0$

$\Rightarrow$$a<b<c<d$