Show that the expression $x^{2} + 2 (a + b + c) x + 3 (b c + c a + a b)$ will be a perfect square if $a = b = c .$

Open in App

Solution

Given quadratic expression will be a perfect square if the discriminant of its corresponding equation is zero.
i.e. $4 {(a + b + c)}^{2} - 4 \times 3 (b c + c a + a b) = 0$
or ${(a + b + c)}^{2} - 3 (b c + c a + a b) = 0$
or $\frac{1}{2} ({(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}) = 0$
which is possible only when $a - b = b - c = c - a = 0$ because square of any number is greater than or equal to $0$

$\Rightarrow a = b = c$

Suggest Corrections

Similar questions

Q. If the expression

x^{2} + 2 (a + b + c) x + 3 (b c + c a + a b)

is a perfect square, then

if a:b:c=1:m:n, where a,b,c are real numbers, and the expressions $x^{2} + 2 (a + b + c) x + 3 (b c + c a + a b)$ is a perfect square, then the value of m+n is

Q. lf

a (b - c) x^{2} + b (c - a) x + c (a - b)

is a perfect square, then

a, b, c

are in

Q. lf

a (b - c) x^{2} + b (c - a) x y + c (a - b) y^{2}

is a perfect square, then

a, b, c

are in

Q. If the equations

x^{2} + c x + a b = 0

and

x^{2} + b x + c a = 0

have a common root, show that

a + b + c = 0

also show that their other roots are and their other root are given by the equation

x^{2} + a x + b c = 0

(where

b \neq c

Join BYJU'S Learning Program

Show that the expression x2+2(a+b+c)x+3(bc+ca+ab) will be a perfect square if a=b=c.

Show that the expression $x^{2} + 2 (a + b + c) x + 3 (b c + c a + a b)$ will be a perfect square if $a = b = c .$