# Proving an equation.

• August 9th 2010, 04:16 AM
Ineedsomehelp
Proving an equation.
1) Prove that if PQ=C and P+Q=B then:

a) X² + bx + c = (x+p)(x+q)
b) X² - bx - c = (x-p)(x+q)

And after answering it can you please explain the procedure :/ Thank you :)
• August 9th 2010, 04:23 AM
HallsofIvy
Quote:

Originally Posted by Ineedsomehelp
1) Prove that if PQ=C and P+Q=B then:

a) X² + bx + c = (x+p)(x+q)
b) X² - bx - c = (x-p)(x+q)

And after answering it can you please explain the procedure :/ Thank you :)

Just do the multiplication on the right. By the "distributive" property, (x+ p)(x+ q)= (x+ p)x+ (x+ p)q. Using the "distributive" property on each of those, $(x+ p)(x+ q)= x(x)+ px+ xq+ pq= x^2+ (p+q)x+ pq$. Since p+ q= B and pq= C, $(x+ p)(x+ q)= x^2+ (p+q)x+ pq= x^2+ Bx+ C$.

You try the second one.

By the way, it is a bad idea to use capital letters and small letters to mean the same thing "P" and "p" are different symbols and imply possibly different values.
• August 9th 2010, 04:29 AM
Ineedsomehelp
:/ I don't totally get it. Just wondering if you could go in a little bit of further explanation off course if you don't mind :) Thank you.
• August 9th 2010, 05:10 AM
earboth
Quote:

Originally Posted by HallsofIvy
Just do the multiplication on the right.
$(x+ p)(x+ q)= (x+ p)x+ (x+ p)q$.
$(x+ p)(x+ q)= x(x)+ px+ xq+ pq= x^2+ (p+q)x+ pq$. Since p+ q= B and pq= C, $(x+ p)(x+ q)= x^2+ (p+q)x+ pq= x^2+ Bx+ C$.

...

Quote:

Originally Posted by Ineedsomehelp
:/ I don't totally get it. <=== are you sure?
Just wondering if you could go in a little bit of further explanation off course if you don't mind :) Thank you.

HallsofIvy took the RHS of the equation and transformed it into the LHs of your equation by comparison of the coefficients.

Of course you can factorize the LHS and show that you have reached the RHS:

$\begin{array}{c}x^2+bx +c &=& x^2+bx +\frac14 b^2 - \frac14b^2 + c \\ =(x+\frac12b)^2-(\frac14b^2-c) \\ &=& (x+\frac12b)^2-\left(\sqrt{\frac14b^2-c}\right)^2 \end{array}$

This is a difference of squares which can be factored to:

$(x+\frac12b)^2-\left(\sqrt{\frac14b^2-c}\right)^2 = \left(x+\underbrace{\frac12b + \sqrt{\frac14b^2-c}}_{p} \right) \left( x+\underbrace{\frac12b - \sqrt{\frac14b^2-c}}_{q} \right)$
• August 9th 2010, 05:14 AM
Ineedsomehelp
ohh thanks well I pretty much get it now ^^ Ty for helping :)