Weird quadratic functions

[thelazyman]

Hi, dont know really where to start on these type of questions, I think there may be a specific formula, but totally forgot it. The question is to find the solutions of the following equations, I tried using quadratic equation, but to no avail, did not work. P and Q are positive paremeters.

the 2 formulas are:

x^2 - 3px + 2p^2 = 0

and x^2 (p + Q)x + pq

Somoneone please help.

 

[skeeter]

x^2 - 3px + 2p^2 = 0
factor ...
(x - 2p)(x - p) = 0
solve ...
x = 2p, x = p


if, for the second one, you meant ...

x^2 + (p + q)x + pq = 0

... that will factor

(x + p)(x + q) = 0

solve as shown previously

 

[stapel]

Somoneone please help.

What were the instructions?

Also, are "Q" and "q" the same variable? Are "p" and "P" the same variable? Is the second expression really an equation?

Are the expression and equation meant to be manipulated together, or are these two separate exercises?

Thank you.

Eliz.

 

[soroban]

Hello, thelazyman!

I assume we are given quadratic equations to solve (not "formulas").

And you do know how to solve a quadratic, don't you?
. . (1) by factoring, or (2) the Quadratic Formula

\(\displaystyle x^2\,-\,3px\,+\,2p^2\:=\:0\)

Factor: \(\displaystyle \x\,-\,p)(x\,-\,2p)\:=\:0\)

Then: \(\displaystyle \,x\,-\,p\:=\:0\;\;\Rightarrow\;\;x\,=\,p\)
And: \(\displaystyle \;\,x\,-\,2p\:=\:0\;\;\Rightarrow\;\;x\,=\,2p\)

\(\displaystyle x^2\,-\,(p + q)x\,+\,pq\:=\:0\)

Factor: \(\displaystyle \x\,-\,p)(x\,-\,q)\:=\:0\)

Then: \(\displaystyle \,x\,-\,p\:=\:0\;\;\Rightarrow\;\;x\,=\,p\)
And: \(\displaystyle \:\,x\,-\,q\:=\:0\;\;\Rightarrow\;\;x\,=\,q\)