Vieta's formulas?

Question

The equation x^2 + px + q = 0 has roots x1 and x2. Find x1^2 + x2^2 as an expression containing p and q.

The equation x^2 + px + q = 0 and has roots x1, and x2. Find (x1-x2)^2 as an expression containing p and q.

If you can't answer both, plz at least answer 1. Thank u.

Answer 1

Viete's formulas.

The equation x^2 + px + q = 0 has roots x1 and x2.
Find x1^2 + x2^2 as an expression containing p and q.

First solve the quadratic.

x1 = [-p + √(p² - 4q)] / 2
x2 = [-p - √(p² - 4q)] / 2

(x1)² + (x2)² = {[-p + √(p² - 4q)] / 2}² + {[-p - √(p² - 4q)] / 2}²

= {[p² - 2p√(p² - 4q) + (p² - 4q)] / 4}
+ {[p² + 2p√(p² - 4q) + (p² - 4q)] / 4}

= {[2p² - 2p√(p² - 4q) - 4q] / 4} + {[2p² + 2p√(p² - 4q) - 4q)] / 4}

= [4p² - 8q] / 4 = p² - 2q
_______________

The equation x^2 + px + q = 0 and has roots x1, and x2. Find (x1 - x2)^2 as an expression containing p and q.

The roots are the same as in the first question.

x1 = [-p + √(p² - 4q)] / 2
x2 = [-p - √(p² - 4q)] / 2

[(x1) - (x2)]² = ({[-p + √(p² - 4q)] / 2} - {[-p - √(p² - 4q)] / 2})²

= (√(p² - 4q)² = p² - 4q

Answer 2

(x-x1)(x-x2) = 0

x^2 - x(x1+x2) - x1 x2 = 0
so p = -(x1+x2) and q = x1 x2

x1^2+x2^2 = (x1+x2)^2 - 2x1 x2 = p^2 - 2q
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Likewise
(x1-x2)^2 = x1^2 - 2x1 x2 + x2^2 = (x1^2 + x2^2) - 2x1 x2
= p^2 - 2q - 2q
= p^2 - 4q

Answer 3

Viete not Vieta
use that x_ 1^2+x_2^2 = (x_1+x_2)^2 -2x_1x_2