Question

Solve the quadratic equation by using quadratic formula $9x^2-3(a+b)x+ab=0$

• A. $x=\frac{a}{3},\frac{b}{3}$
• B. $x=\frac{-a}{3},\frac{-b}{3}$
• C. $x=\frac{-a}{3},\frac{b}{3}$
• D. $x=\frac{a}{3},\frac{-b}{3}$

Answer 1

Answer: (A)

$x=\frac{a}{3},\frac{b}{3}$

$\Rightarrow$ Here, $a=9,b=-3(a+b),c=ab$

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

$=\frac{-[-3(a+b\left)\right]\pm \sqrt{[-3(a+b){]}^2-4(9\left)\right(ab)}}{2\left(9\right)}$

$=\frac{3a+3b\pm \sqrt{9(a^2+b^2+2ab)-36ab}}{18}$

$=\frac{3a+3b\pm \sqrt{9a^2+9b^2+18ab-36ab}}{18}$

$=\frac{3a+3b\pm \sqrt{9a^2+9b^2-18ab}}{18}$

$\frac{3a+3b\pm \sqrt{9(a^2+b^2-2ab)}}{18}$

$=\frac{3a+3b\pm \sqrt{9(a-b{)}^2}}{18}$

$=\frac{3a+3b\pm 3(a-b)}{18}$

$=\frac{3a+3b\pm 3a-3b}{18}$

$x=\frac{3a+3b+3a-3b}{18}$ and $x=\frac{3a+3b-3a+3b}{18}$

$x=\frac{6a}{18}$ and $x=\frac{6b}{18}$

$\therefore$ $x=\frac{a}{3},\frac{b}{3}$