An algebra problem by Razif FA

Algebra Level pending

$\large x^2 - (3p-2) x + (2p+8) = 0$

Given that $x_1$ and $x_2$ are the roots of the quadratic equation above.

If $p$ is positive and $x_1, p, x_2$ formed a geometric progression , compute $x_1 + p + x_2$ .

15 11 13 10 12 14

1 solution

Hung Woei Neoh
Jun 13, 2016

$x^2-(3p-2)x+(2p+8) = 0$

Given that the roots are $x_1$ and $x_2$ . From Vieta's formula,

$x_1+x_2 = 3p-2\\ x_1x_2=2p+8$

Now, we know that $x_1,p,x_2$ forms a GP, therefore,

$r=\dfrac{p}{x_1}=\dfrac{x_2}{p}\\ \implies p^2 = x_1x_2$

Substitute this into the product of roots:

$p^2 = 2p+8\\ p^2-2p-8=0\\ (p-4)(p+2)=0\\ p=4,\;-2$

It is given that $p>0$ , therefore, $p=4$ . From sum of roots,

$x_1+x_2 = 3p-2\\ x_1+p+x_2 = 4p-2 = 4(4)-2 = \boxed{14}$