Question

Hard algebra question! Please help?

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

Answer 1

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

Explanation:

Have a look:

Answer 2

`(3/2) * 2 = 3` and `(-4/2)^2 = 4` thus,
`2p+2q = 3` and `p^2q^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 = -b/a`, `pq = c/a`
proof:
`a(x-r_1\)(x-r_2)=ax^2+bx+c`
`ax^2 - a(r_1 + r_2)x + a(r_1)(r_2)= ax^2+bx+c`
Thus `r_1 + r_2 = -b/a and (r_1)(r_2) = c/a`
`p + q = -3/2, pq = 4/2 = 2`

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

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

`p+q = \frac{3+\sqrt{41}+3-\sqrt{41}}{4} = \frac{6}{4}=3/2`
`pq = \frac{-32}{16}= -2`
`2(p+q) = 3 and p^2q^2 = 4`