Quadratics (year 12 high school) question

Find the numerical value of k when the equation 2x^2-(k+2)x-k^2=0, the square of the sum of the roots exceeds the product by 2.

I know how to solve a roots question, but what do I do with the "exceeds bit", I don't quite understand.

Thank-you for any help!

Roots : a and b
a + b = (k + 2)/2
ab = -k^2/2

(a + b)^2 = ab + 2
((k + 2)/2)^2 = (-k^2/2) + 2
(k + 2)^2 = -2k^2 + 8
k^2 + 4k + 4 = -2k^2 + 8
3k^2 + 4k - 4 = 0
(3k - 2)(k + 2) = 0
k = -2, 2/3

You're welcome :)

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2x² - (k+2)x - k² = 0.
By the quadratic formula,
x = ( (k+2) ± √((k+2)² - 4(2)(-k)) ) / 2(2)
= ( (k+2) ± √((k+2)² + 8k) ) / 4
= ( (k+2) ± √(k² + 12k + 4) ) / 4.

So x = ((k+2) + √(k² + 12k + 4)) / 4
or x = ((k+2) - √(k² + 12k + 4)) / 4.

The sum of these roots is (k+2)/4; the square of this sum is (k+2)²/16.

The product of these roots is:
(k+2)² + (k+2)(-√(k²+12k+4)) + (k+2)(√(k²+12k+4)) - (√(k²+12k+4))²
= (k+2)² - (k²+12k+4)
= k² + 4k + 4 - k² - 12k - 4
= -8k.

If the square of the sum of the roots exceeds their product by 2, then
(k+2)²/16 = -8k + 2.

Simplify:
(k+2)² = -128k + 32, or
k² + 4k + 4 = -128k + 32, or
k² + 132k - 28 = 0.

Using the quadratic formula again, we have
k = ( -132 ± √(132² - 4(1)(-28) ) / 2
= (-132 ± √17536) / 2
= (-132 ± √(64·274)) / 2
= (-132 ± 8√274) / 2
= -66 ± 4√274.

The approximate values for k are 0.21178 and -132.21178.

in words:

"the square of the sum of the roots exceeds the product by 2."

in mathematical expression

(x₁ + x₂)² = (x₁)(x₂) + 2