Question

If p + q = 12 and pq = 14, then p²+ q²
please answer it fast​

Answer 1

$\huge\mathtt{Hello!}$

$p + q = 12 \: \: \: \: \: pq = 14$

${(p + q)}^{2} = {p}^{2} + {q}^{2} +2pq$

${p}^{2} + {q}^{2} = {(p + q)}^{2} - 2pq$

$= {(12)}^{2} - 2(14)$

$= 144 - 28$

${p}^{2} + {q}^{2} = 116$

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Answer 2

Step-by-step explanation:

this example, we’ll choose to solve for p.

p + q = 3

Subtract q from both sides:

p = 3-q

Now substitute for p in the second equation:

p^2 - q^2 = 15

p = 3-q

(3-q)^2 - q^2 = 15

Expand the parentheses on the left, and then use the FOIL method to distribute the terms:

(3-q)(3-q)-q^2 = 15

9 - 6q + q^2 - q^2= 15

Handily, both of the q^2’s cancel out:

9–6q = 15

Subtract 9 from both sides:

-6q = 6

Divide by -6:

q = -1

If q = -1, then if p = 3-q:

p = 3-q

q = -1

p = 3 - (-1)

p = 3 +1

p = 4

So p = 4 and q = -1. Let’s double-check these work by substituting for p and q in both of the original equations.

p + q = 3

p = 4

q = -1

4 + (-1) = 3

4 - 1 = 3

3 = 3

p^2 - q^2 = 15

p = 4

q = -1

4^4 - (-1)^2 = 15

16 - 1 = 15

15 = 15

So p = 4 and q = -1.