I am having trouble with a problem in an old maths textbook, as follows:
Question: Prove that the roots of the equation:
[math](p - q - r)x^2 + px + q + r = 0[/math] are real if p, q, and r are real.
Solution: The condition for real roots is [math](b^2 ≥ 4ac)[/math] is here that
Comment: I understand the solution but I am having trouble with the steps of simplifying [math]p^2 ≥ 4(p-q-r)(q+r)[/math] to [math]p^2 - 4p(q+r) +4)(q+r)^2 ≥ 0[/math] to the form [math](p-2(q+r))^2 ≥ 0[/math]
Question: Prove that the roots of the equation:
[math](p - q - r)x^2 + px + q + r = 0[/math] are real if p, q, and r are real.
Solution: The condition for real roots is [math](b^2 ≥ 4ac)[/math] is here that
[math]p^2 ≥ 4(p-q-r)(q+r)[/math]
i.e. that [math]p^2 - 4p(q+r) +4)(q+r)^2 ≥ 0[/math]
or [math](p-2(q+r))^2 ≥ 0[/math]
This is always true for the left hand side is the square of a real quantity and therefore cannot be negative.
Comment: I understand the solution but I am having trouble with the steps of simplifying [math]p^2 ≥ 4(p-q-r)(q+r)[/math] to [math]p^2 - 4p(q+r) +4)(q+r)^2 ≥ 0[/math] to the form [math](p-2(q+r))^2 ≥ 0[/math]
Could you please let me know the steps to do this. Many thanks in advance.