# Find the equation of parabola whose intercepts on x-axis are -2+sqrt(-5) and -2-sqrt(-5)?

Feb 26, 2017

You can only have real intercepts on $x$-axis and not complex. But still one could try it the following way, but we may not have the desired intercepts in such case. The equation is $y = \frac{5}{9} \left({x}^{2} + 4 x + 9\right)$.

#### Explanation:

You can only have real intercepts on $x$-axis and not complex. But still one could try it the following way, but we may not have the desired intercepts in such case.

The standard form of an equation with the $x$-intercepts $\alpha$ and $\beta$ is $y = a \left(x - \alpha\right) \left(x - \beta\right)$

Hence an equation with the $x$-intercepts as $- 2 + \sqrt{- 5}$ and $- 2 - \sqrt{- 5}$ is

$y = a \left(x + 2 + \sqrt{- 5}\right) \left(x + 2 - \sqrt{- 5}\right)$

= $a \left({x}^{2} + 4 x + 4 - \left(- 5\right)\right)$

= $a \left({x}^{2} + 4 x + 9\right)$

As it passes through $\left(- 4 , 5\right)$, we have

$5 = a \left({\left(- 4\right)}^{2} + 4 \left(- 4\right) + 9\right) = a \left(16 - 16 + 9\right) = 9 a$

i.e. $a = \frac{5}{9}$

and equation is $y = \frac{5}{9} \left({x}^{2} + 4 x + 9\right)$