Question

If $a,b,c,d$ are four consecutive terms of an increasing A.P., then the roots of the equation
$(x-a)(x-c)+2(x-b)(x-d)=0$ are

• A. $\text{real and distinct}$
• B. $\text{non-real complex}$
• C. $\text{real and equal}$
• D. $\text{integers}$

Answer 1

The correct option is A $\text{real and distinct}$

$a,b,c,d$ are $4$ consecutive terms of A

Let $a=m-3n,b=m-n.c=m+n,d=m+3n$

${}^{\prime }2n^{\prime }$ is common difference

$(x-a)(x-c)+2(x-b)(x-d)=0$

$3x^2-(a+c+2b+2d)x+(ac+2bd)=0$

$=6m+2n$

$ac+2bd=m^2-2mn-3n^2+2m^2+4mn-6n^2$

$=3m^2+2mn-9n^2$

$3x^2-(6m+2n)x+(3m^2+2mn-9n^2)=0$

$△={(6m+2n)}^2-4\left(3\right)(3m^2+2mn-9n^2)$

$=4(9m^2+6mn+n^2-9m^2-6mn+36n^2)$

$=4\left(37\right)n^2$

$△>0$

$\therefore$ Roots of $(x-a)(x-c)+2(x-b)(x-d)$ are real and distinct.