Question

The origin and the roots of the equation $x^2+ax+b$ = 0 form an equilateral triangle, if :

• A. a²=b
• B. b²=3a
• C. b²=a
• D. a²=3b

Answer 1

The correct option is D

a²=3b

$x^2+ax+b$ = 0 $\Rightarrow \frac{-a\pm \sqrt{a^2-4b}}{2}$

If $a^2-4b\le 0$ then the three points, 0 and the two roots are real and so collinear. Therefore, considering $a^2-4b<0$, we have roots as $\frac{-a\pm \sqrt{4b-a^2}}{2}$

Let A = $-\frac{a}{2}+i\frac{\sqrt{4b-a^2}}{2}$, B = $-\frac{a}{2}-i\frac{\sqrt{4b-a^2}}{2}$

OA = $\sqrt{\frac{a^2}{4}+\frac{4b-a^2}{4}}$ = $\sqrt{b}$ = OB

AB = $\sqrt{(4b-a^2)}$. Since OA = OB = AB,

we get $4b-a^2$ = b i.e., $a^2$ = $3b$