Question

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

• A. Real and distinct
• B. Real and equal
• C. Imaginary
• D. one real and one imaginary

Answer 1

The correct option is A

Real and distinct

Given equation can be written as

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

Its Discriminant D is

$=(a+c+2b+2d{)}^2-4.3(ac+2bd)$

$=\left\{\right(a+2d)+(c+2b){\}}^2-12(ac+2bd)$

$=\left\{\right(a+2d)-(c+2b){\}}^2+4(a+2d\left)\right(c+2b)-12(ac+2bd)$

$=\left\{\right(a+2d)-(c+2b){\}}^2-8ac+8ab+8dc-8bd$

$=\left\{\right(a+2d)-(c+2b){\}}^2+8(c-b\left)\right(d-a)$

which is +ve since a < b < c < d. Hence roots are real and distinct.