• May 20th 2010, 02:36 PM
Hello everyone, I need help with one more question and that's using the quadractic formula to solve 16x2 + 96x + 128 = 0 and -4x2 + 28x – 48 = 0. Could someone tell me the answer to both so I can hurry and submit the assignment, I've been really waiting to see what I get on this.
• May 20th 2010, 02:49 PM
pickslides
But if we solve these equations for your assignment it wont be your work.

I suggest studying this, for $\displaystyle ax^2+bx+c=0$ then $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

It will give you the answer.
• May 20th 2010, 02:53 PM
Quote:

Originally Posted by pickslides
But if we solve these equations for your assignment it wont be your work.

I suggest studying this, for $\displaystyle ax^2+bx+c=0$ then $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

It will give you the answer.

I know, I've been studying this formula and I have been getting all kinds of things, that's why I posted the question.
• May 20th 2010, 03:05 PM
skeeter
Quote:

Hello everyone, I need help with one more question and that's using the quadractic formula to solve 16x2 + 96x + 128 = 0 and -4x2 + 28x – 48 = 0. Could someone tell me the answer to both so I can hurry and submit the assignment, I've been really waiting to see what I get on this.

$\displaystyle 16x^2 + 96x + 128 = 0$

divide every term by 16 ....

$\displaystyle x^2 + 6x + 8 = 0$

this is a factorable quadratic ... do it.

---------------------------------------

$\displaystyle -4x^2 + 28x - 48 = 0$

start by dividing every term by -4 ... same song, second verse.
• May 20th 2010, 03:09 PM
spruancejr
Break down the first term:
$\displaystyle 16x^2$ compared to $\displaystyle ax^2$

Do you see to what the "a" term may be equal?
• May 21st 2010, 04:01 AM
mr fantastic
Quote: