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:

$$(p - q - r)x^2 + px + q + r = 0$$ are real if p, q, and r are real.

Solution: The condition for real roots is $$(b^2 ≥ 4ac)$$ is here that

$$p^2 ≥ 4(p-q-r)(q+r)$$​

i.e. that $$p^2 - 4p(q+r) +4)(q+r)^2 ≥ 0$$​

or $$(p-2(q+r))^2 ≥ 0$$​

​

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 $$p^2 ≥ 4(p-q-r)(q+r)$$ to $$p^2 - 4p(q+r) +4)(q+r)^2 ≥ 0$$ to the form $$(p-2(q+r))^2 ≥ 0$$

​

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 $p^2-4p(q+r) + 4 \mathbf{)}(q+r)^2$ .
After that, try expanding $(p-2(q+r))^2$ and see what you get.

 

[skeeter]

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

$p^2 \ge 4p(q+r)-4(q+r)^2$

$p^2 -4p(q+r) +4(q+r)^2 \ge 0$

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

 

[Mith]

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

Thank you for pointing that out!

 

[Mith]

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

$p^2 \ge 4p(q+r)-4(q+r)^2$

$p^2 -4p(q+r) +4(q+r)^2 \ge 0$

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

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