Equation x^2 + px - q = 0 has roots alpha, beta. Equation x^2 - p'x + q' = 0 has roots alpha', beta'. Find the value of (alpha - alpha')^2 + (beta - alpha')^2 + (alpha - beta')^2 + (beta - beta')^2

We have the equations 

x^2 + px - q = 0     (equation 1)    Roots: α, β

x^2 - p'x + q' = 0    (equation 2)    Roots: α', β'

Using the sum of roots and product of roots formulae for equation 1, we have

α + β = -(p)/1 = -p

αβ = (-q)/1 = -q

Using the same on equation 2, 

α' + β' = -(-p')/1 = p'

α'β' = q'/1 = q'

Now, lets simplify (α - α')² + (β - α')² + (α - β')² + (β - β')² to something useful. Basically, we'll try to transform this into a state where we can easily use the results we found above.

= (α - α')² + (β - α')² + (α - β')² + (β - β')²

= (α² + α²- 2αα') + (β² + α'² - 2βα') + (α² + β'² - 2αβ') + (β² + β'² - 2ββ')

= 2α² + 2α'² + 2β² +2β'² - 2αα' - 2βα' - 2αβ' - 2ββ'

= 2(α² + β²) + 2(α'² + β'²) - 2α'(α + β) - 2β'(α + β)

We'll now use a² + b² = (a + b)² - 2ab in the first two terms. The other two terms will be simplified by taking (α + β) common.

= 2[(α + β)² - 2αβ] + 2[(α' + β')² - 2α'β'] - 2(α + β)(α' + β')

This is what we needed. The four results we found in the beginning can now be used.

= 2[(-p)² - 2(-q)] + 2[(p')² - 2(q')] - 2(-p)(p')

= 2(p² + 2q) + 2(p'² - 2q') + 2pp'

= 2p² + 4q + 2p'² - 4q' + 2pp'

= 2(p² - p'²) + 4(q - q') + 2pp'

2(p² - p'²) + 4(q - q') + 2pp' is the answer.