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

2 Answers
May 18, 2018

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

Explanation:

Have a look:

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May 18, 2018

(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