Question

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

• A. real and distinct
• B. non real complex
• C. real and equal
• D. integer

Answer 1

Answer: (A)

real and distinct

Since, $a,b,c,d$ are in A.P.
Let $a=m-n$, $b=m$, $c=m+n$, $d=m+2n$,
where, $n\ne 0$
$(x-a)(x-c)+2(x-b)(x-d)=0$
$\Rightarrow 3x^2-x(a+c+2b+2d)+ac+2bd=0$
$\Rightarrow 3x^2-x(m-n+m+n+2m+2m+2n)+(m-n)(m+n)+2\left(m\right)(m+2n)=0$
$\Rightarrow 3x^2-2x(3m+2n)+3m^2+4mn-n^2=0$
Now, $D=b^2-4ac=4[{(3m+2n)}^2-3(3m^2+4mn-n^2)]=28n^2>0$
Therefore, roots are real and distinct.
Ans: A