How do I solve this ?

Question

Given p-2 is one of the roots of the quadratic equation $x^{2}$ $x^2$ - m $x$ $x$ + 9 $- p^{2}$ $-p^2$ = 0 . Express p in terms of m.

1430 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-27T12:00:10

$p = \frac{2 m + 13}{m + 4}$ $p = frac(2 m + 13)(m + 4)$

$p - 2$ $p - 2$ is a root of $f (x) = x^{2} - m x + 9 - p^{2}$ $f(x) = x^(2) - m x + 9 - p^(2)$ .

So ${(p - 2)}^{2} - m (p - 2) + 9 - p^{2} = 0$ $(p - 2)^(2) - m (p - 2) + 9 - p^(2) = 0$ .

Let's expand the parentheses:

$\Rightarrow p^{2} - 4 p + 4 - m p + 2 m + 9 - p^{2} = 0$ $Rightarrow p^(2) - 4 p + 4 - m p + 2 m + 9 - p^(2) = 0$

Then, let's simplify the equation:

$\Rightarrow p^{2} - p^{2} - 4 p - m p + 2 m + 9 + 4 = 0$ $Rightarrow p^(2) - p^(2) - 4 p - m p + 2 m + 9 + 4 = 0$

$\Rightarrow - p (4 + m) + 2 m + 13 = 0$ $Rightarrow - p (4 + m) + 2 m + 13 = 0$

Now, let's solve for $p$ $p$ :

$\Rightarrow - p (4 + m) = - (2 m + 13)$ $Rightarrow - p (4 + m) = - (2 m + 13)$

$\Rightarrow - p = - \frac{2 m + 13}{4 + m}$ $Rightarrow - p = - frac(2 m + 13)(4 + m)$

$therefore p = frac(2 m + 13)(m + 4)$