First slide

Theory of equations

Question

If the expression $x^{2} + 2 (a + b + c) x + 3 (bc + ca + ab)$ is a perfect square, then

Moderate

A

a = b = c
B

$a = \pm b = \pm c$
C

$a = b \neq c$
D

none of these

Solution

Given quadratic expression is $x^{2} + 2 (a + b + c) x + 3 (bc + ca + ab)$ .
This quadratic expression will be a perfect square if the
discriminant of its corresponding equation is zero. Hence,
$4 (a + b + c)^{2} - 4 \times 3 (bc + ca + ab) = 0$
or    $(a + b + c)^{2} - 3 (bc + ca + ab) = 0$
or    $a^{2} + b + c^{2} + 2 ab + 2 bc + 2 ca - 3 (bc + ca + ab) = 0$
or    $a^{2} + b^{2} + c^{2} - abbc - ca = 0$
or    $\frac{1}{2} [2 a^{2} + 2 b^{2} + 2 c^{2} - 2 ab - 2 bc - 2 ca] = 0$
or    $\frac{1}{2} [(a^{2} + b^{2} - 2 ab) + (b^{2} + c^{2} - 2 bc) + (c^{2} + a^{2} - 2 ca)] = 0$
or $\frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0$
which is possible only when $(a - b)^{2} = 0, (b - c)^{2} = 0$
and $(c - a)^{2} = 0, i.e., a = b = c$

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