# Given that one root is 3 times the others for the quadratic equation 3x^2-2x+p=0, find (a) the value of p and (b) the two roots ?

May 4, 2018

$p = \frac{1}{4} , x = \frac{1}{6} , \frac{1}{2}$

#### Explanation:

Let the roots be $r , 3 r$:

$a \left(x - r\right) \left(x - 3 r\right) = 0$

$a {x}^{2} - 4 a r x + 3 a {r}^{2} = 0$

$3 {x}^{2} - 2 x + p = a {x}^{2} - 4 a r x + 3 a {r}^{2}$

$a = 3$

$3 {x}^{2} - 2 x + p = 3 {x}^{2} - 12 r x + 9 {r}^{2}$

$12 r = 2$

$r = \frac{1}{6} \implies$ $3 r = \frac{1}{2}$

$p = 9 \cdot \frac{1}{36} = \frac{1}{4}$

Check:

$3 {x}^{2} - 2 x + \frac{1}{4} = 0$
$3 \left({x}^{2} - \left(\frac{2}{3}\right) x\right) = - \frac{1}{4}$
$3 \left({x}^{2} \left(- \frac{2}{3}\right) x + {\left(\frac{1}{3}\right)}^{2}\right) = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$
${\left(x - \frac{1}{3}\right)}^{2} = \frac{1}{36}$
$x - \frac{1}{3} = \pm \frac{1}{6}$
$x = \frac{1}{3} \pm \frac{1}{6}$
$x = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6}$
$x = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$