If a^2 - 6b^2 - ab = 0, what is value of b/a ?

Aug 14, 2018

$- \frac{1}{2} , \mathmr{and} , \frac{1}{3}$.

Explanation:

Given that, ${a}^{2} - 6 {b}^{2} - a b = 0$.

$\therefore 6 {b}^{2} + a b - {a}^{2} = 0$.

$\therefore \underline{6 {b}^{2} + 3 a b} - \underline{2 a b - {a}^{2}} = 0$.

$\therefore 3 b \left(2 b + a\right) - a \left(2 b + a\right) = 0$.

$\therefore \left(2 b + a\right) \left(3 b - a\right) = 0$.

$\therefore 2 b + a = 0 , \mathmr{and} , 3 b - a = 0$.

$\therefore 2 b = - a , \mathmr{and} , 3 b = a$.

$\therefore \frac{b}{a} = - \frac{1}{2} , \mathmr{and} , \frac{b}{a} = \frac{1}{3}$.

Aug 14, 2018

$\frac{b}{a} = \frac{1}{3} \mathmr{and} \frac{b}{a} = - \frac{1}{2}$

Explanation:

Here ,

${a}^{2} - 6 {b}^{2} - a b = 0$

Dividing each term by ${a}^{2} \ne 0$

$1 - 6 {b}^{2} / {a}^{2} - \frac{b}{a} = 0$

$\implies 6 {\left(\frac{b}{a}\right)}^{2} + \frac{b}{a} - 1 = 0$

For simplicity take $\frac{b}{a} = x$

$\therefore 6 {x}^{2} + x - 1 = 0$

$\implies 6 {x}^{2} + 3 x - 2 x - 1 = 0$

$\implies 3 x \left(2 x + 1\right) - 1 \left(2 x + 1\right) = 0$

$\implies \left(3 x - 1\right) \left(2 x + 1\right) = 0$

$\implies 3 x - 1 = 0 \mathmr{and} 2 x + 1 = 0$

$\implies x = \frac{1}{3} \mathmr{and} x = - \frac{1}{2}$

Subst. back $x = \frac{b}{a}$

$\therefore \frac{b}{a} = \frac{1}{3} \mathmr{and} \frac{b}{a} = - \frac{1}{2}$