Difficulty simplifying terms in proof

[Mith]

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

[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.​

 

[blamocur]

First, you want to get rid of a stray right parenthesis after 4 in [imath]p^2-4p(q+r) + 4 \mathbf{)}(q+r)^2[/imath] .
After that, try expanding [imath](p-2(q+r))^2[/imath] and see what you get.

 

[skeeter]

[imath]p^2 \ge [4p-4(q+r)](q+r)[/imath]

[imath]p^2 \ge 4p(q+r)-4(q+r)^2[/imath]

[imath]p^2 -4p(q+r) +4(q+r)^2 \ge 0[/imath]

[imath][p - 2(q+r)]^2 \ge 0[/imath]

 

[Mith]

First, you want to get rid of a stray right parenthesis after 4 in [imath]p^2-4p(q+r) + 4 \mathbf{)}(q+r)^2[/imath] .
After that, try expanding [imath](p-2(q+r))^2[/imath] and see what you get.

Thank you for pointing that out!

 

[Mith]

[imath]p^2 \ge [4p-4(q+r)](q+r)[/imath]

[imath]p^2 \ge 4p(q+r)-4(q+r)^2[/imath]

[imath]p^2 -4p(q+r) +4(q+r)^2 \ge 0[/imath]

[imath][p - 2(q+r)]^2 \ge 0[/imath]

Thank you! That is really helpful. I now see that -1 is actually a common factor for [imath]q[/imath] and [imath]r[/imath] in the first set of brackets which allows [imath]4(p−q−r)[/imath] to be written as [imath][4p−4(q+r)][/imath]. Great !