Question

Solve the following quadratic equation using quadratic formula .

$9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0$

• A. The roots are $\frac{2a+b}{3}$ and $\frac{a+2b}{3}$
• B. The roots are $\frac{2a+b}{3}$ and $\frac{a-2b}{3}$
• C. The roots are $\frac{5a+b}{3}$ and $\frac{a+2b}{3}$
• D. The roots are $\frac{2a+b}{3}$ and $\frac{a-2b}{4}$

Answer 1

The correct option is A

The roots are $\frac{2a+b}{3}$ and $\frac{a+2b}{3}$

We have,

$9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0$

Comparing this equation with $A{x}^2+Bx+C=0$, we have

$A=9,B=-9(a+b)andC=2a^2+5ab+2b^2$

$\therefore D=B^2-4AC$

$\Rightarrow D=81(a+b{)}^2-36(2a^2+5ab+2b^2)$

$\Rightarrow D=81(a^2+b^2+2ab)-(72a^2+180ab+72b^2)$

$\Rightarrow D=9a^2+9b^2-18ab$

$\Rightarrow D=9(a^2+b^2-2ab)$

$\Rightarrow D=9(a-b{)}^2\ge 0$

$\Rightarrow D\ge 0$

So, the roots of the given equation are real and are given by

$\alpha =\frac{-B+\sqrt{D}}{2A}=\frac{9(a+b)+3(a-b)}{18}=\frac{12a+6b}{18}=\frac{2a+b}{3}$

and, $\beta =\frac{-B-\sqrt{D}}{2A}=\frac{9(a+b)-3(a-b)}{18}=\frac{6a+12b}{18}=\frac{a+2b}{3}$