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 :)

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- Aug 9th 2010, 04:16 AMIneedsomehelpProving 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 :) - Aug 9th 2010, 04:23 AMHallsofIvy
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, $\displaystyle (x+ p)(x+ q)= x(x)+ px+ xq+ pq= x^2+ (p+q)x+ pq$. Since p+ q= B and pq= C, $\displaystyle (x+ p)(x+ q)= x^2+ (p+q)x+ pq= x^2+ Bx+ C$.

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. - Aug 9th 2010, 04:29 AMIneedsomehelp
:/ 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.

- Aug 9th 2010, 05:10 AMearboth
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:

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

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

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