## If $2 {p}^{2} - 3 p - 4 = 0$ and $2 {q}^{2} - 3 q - 4 = 0$ (p does not equal to 0), find the value of $2 p + 2 q$ and ${p}^{2} {q}^{2}$

May 18, 2018

I tried this...the procedure should be ok...BUT check my maths anyway.

#### Explanation:

Have a look:

May 18, 2018

$\left(\frac{3}{2}\right) \cdot 2 = 3$ and ${\left(- \frac{4}{2}\right)}^{2} = 4$ thus,
$2 p + 2 q = 3$ and ${p}^{2} {q}^{2} = 4$

#### Explanation:

Quick way: You may use Vieta's Formulas
First notice that p and q have the exact same equation and thus will have the same solution,
$p + q = - \frac{b}{a}$, $p q = \frac{c}{a}$
proof:
$a \left(x - {r}_{1} \setminus\right) \left(x - {r}_{2}\right) = a {x}^{2} + b x + c$
$a {x}^{2} - a \left({r}_{1} + {r}_{2}\right) x + a \left({r}_{1}\right) \left({r}_{2}\right) = a {x}^{2} + b x + c$
Thus ${r}_{1} + {r}_{2} = - \frac{b}{a} \mathmr{and} \left({r}_{1}\right) \left({r}_{2}\right) = \frac{c}{a}$
$p + q = - \frac{3}{2} , p q = \frac{4}{2} = 2$

Long way:
solve for $2 {p}^{2} - 3 p - 4 = 0$
$p = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$
Sub in a =2, b = -3 and c = -4
p = \frac{3 \pm \sqrt{9 - 4(2)(-4}}{2(2)}
$p = \setminus \frac{3 \setminus \pm \setminus \sqrt{9 + 32}}{4}$
$p = \setminus \frac{3 \setminus \pm \setminus \sqrt{41}}{4}$

$p = \setminus \frac{3 + \setminus \sqrt{41}}{4}$, $p = \setminus \frac{3 - \setminus \sqrt{41}}{4}$
q has the exact same equation and is thus have the same solution:
$q = \setminus \frac{3 + \setminus \sqrt{41}}{4}$, $q = \setminus \frac{3 - \setminus \sqrt{41}}{4}$

$p + q = \setminus \frac{3 + \setminus \sqrt{41} + 3 - \setminus \sqrt{41}}{4} = \setminus \frac{6}{4} = \frac{3}{2}$
$p q = \setminus \frac{- 32}{16} = - 2$
$2 \left(p + q\right) = 3 \mathmr{and} {p}^{2} {q}^{2} = 4$