# The quadratic equation  3x^2 -9x +b =0  has roots alpha and alpha+2, find b?

Apr 1, 2017

$b = \frac{15}{4}$

#### Explanation:

Suppose the roots of the general quadratic equation:

$a {x}^{2} + b x + c = 0$

are $\alpha$ and $\beta$ , then using the root properties we have:

$\text{sum of roots} \setminus \setminus \setminus \setminus \setminus \setminus = \alpha + \beta = - \frac{b}{a}$
$\text{product of roots} = \alpha \beta \setminus \setminus \setminus \setminus = \frac{c}{a}$

So for the given quadratic with roots $\alpha$ and $\beta$:

$3 {x}^{2} - 9 x + b = 0$

we know that:

$\alpha + \beta = - \frac{- 9}{3} = 3 \setminus \setminus \setminus$ ; and $\setminus \setminus \setminus \alpha \beta = \frac{b}{3}$

But we also know that $\beta = \alpha + 2$

Hence,

$\alpha + \beta = = \alpha + \left(\alpha + 2\right) = 2 \alpha + 2 \implies 2 \alpha + 2 = 3$
$\therefore \alpha = \frac{1}{2}$

and:

$\alpha \beta = = \alpha \left(\alpha + 2\right) = \frac{1}{2} \left(\frac{1}{2} + 2\right) = \frac{5}{4}$
$\therefore \frac{5}{4} = \frac{b}{3} \implies b = \frac{15}{4}$