First slide

Theory of equations

Question

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

Moderate

A

Real and distinct
B

Real and equal
C

Imaginary
D

None of these

Solution

Given equation can be rewritten as
$3 x^{2} - (a + c + 2 b + 2 d) x + (ac + 2 bd) = 0$
Its discriminant D
$= {(a + c + 2 b + 2 d)}^{2} - 4.3 (ac + 2 bd)$
$= {\{(a + 2 d) + (c + 2 b)\}}^{2} - 12 (ac + 2 bd)$
$= {\{(a + 2 d) - (c + 2 b)\}}^{2} - 8 ac + 8 ab + 8 dc - 8 bd$
$= {\{(a + 2 d) - (c + 2 b)\}}^{2} + 8 (c - b) (d - a)$
Which is $+ ve,$ since $a < b < c < d .$ Hence roots are real and distinct.

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