Question

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

• A. a + b + c
• B. 2a + b + c
• C. 3abc
• D. ab + bc + ca

Answer 1

Answer: (D)

Given equation is
$(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a) = 0$
$\Rightarrow \left(x^{2} - bx - ax + ab\right) + \left(x^{2} - cx - bx + bc\right) + \left(x^{2} - ax - cx + ac\right) = 0$
$\Rightarrow \left[x^{2} - \left(a + b\right)x + ab\right] + \left[x^{2} - \left(b + c\right)x + bc\right]+ \left[x^{2 }- \left(a + c\right)x + ac\right] = 0$
$\Rightarrow 8x^{2} - 2\left(a + b + c\right)x + ab + bc + ca = 0$
Since roots are equal
$\therefore \quad$ We have
$b^{2} - 4ac = 0 \Rightarrow b^{2} = 4ac$
$4\left(a + b + c\right)^{2 }= 12\left(ab + bc + ca\right)$
$\left(a + b + c\right)^{2} = 3\left(ab + bc + ca\right)$
$a^{2 }+ b^{2} + c^{2} = ab + bc + ca.$