Question

If $a<b<c<d$, then the roots of the equation
$(x-a)(x-c)+2(x-b)(x-d)=0$ are real and distinct.

• A. True
• B. False

Answer 1

The correct option is A True

(i) The given equation can be written as
$3x^2-(a+c+2b+2d)x+ac+2bd=0$
$\Delta ={(a+c+2b+2d)}^2-12(ac+2bd)$
$={\left[\right(a+2d)-(c+2b\left)\right]}^2+4(a+2d)(c+2b)-12(ac+2bd)$
$={\left[\right(a+2d)-(c+2b\left)\right]}^2+8ab+8cd-8ac-8bd$
$={\left[\right(a+2d)-(c+2b\left)\right]}^2+8(c-b)(d-a)>0$
Since $a<b<c<d$ so that $(c-b)(d-a)>0$.
Hence the roots are real and distinct.