Algebra II: Complex Numbers & Trinomials & Factoring

**Drew_445** · November 11th 2008, 02:17 PM

I need help with these problems please:
Solve by Factoring:
$3x^2 + 6x - 24 = 0$
The problem i got is that any combination of number I got that produced -24 didn't add to 6.

And:

$5x^2 - 40x - 45 = 0$

And:

$9x^2 + 24x + 16$

Solve the equation.
$4x^2 + 9 = 0$
The answers were:
a. $\frac{-3}{2}i, \frac{3}{2}i$ b. $\frac{-9}{4}i , \frac{9}{4}i$

Use the Quadratic Formula to solve the equation.
$-2x^2 - 5x + 3 = 0$
My solution didn't match the 2 possible answers provided:
$\frac{-27}{2} , 11$ and $-3 , \frac{1}{2}$

$4x^2 - x + 9 = 0$
provided possible answers:
$\frac{1}{4} +- \frac{i\sqrt143}{4}$ or $\frac{1}{8} +- \frac{i\sqrt143}{8}$

I know its kind of alot but I absolutely cannot figure this out! Help is very appreciated

**mr fantastic** · November 11th 2008, 02:32 PM

Originally Posted by Drew_445

I need help with these problems please:
Solve by Factoring:
$3x^2 + 6x - 24 = 0$
The problem i got is that any combination of number I got that produced -24 didn't add to 6.

Mr F says: Start by writing it as 3(x^2 + 2x - 8) = 0.

And:

$5x^2 - 40x - 45 = 0$

Mr F says: Start by writing it as 5(x^2 - 8x - 9) = 0.

And:

$9x^2 + 24x + 16$

Mr F says: This has the form of a perfect square.

Solve the equation.
$4x^2 + 9 = 0$

Mr F says: ${\color{red}x^2 = -\frac{9}{4} \Rightarrow x = \sqrt{-\frac{9}{4}} = \pm i \, \frac{3}{2}}$ .

The answers were:
a. $\frac{-3}{2}i, \frac{3}{2}i$ b. $\frac{-9}{4}i , \frac{9}{4}i$

Use the Quadratic Formula to solve the equation.
$-2x^2 - 5x + 3 = 0$
My solution didn't match the 2 possible answers provided:
$\frac{-27}{2} , 11$ and $-3 , \frac{1}{2}$

Mr F says: Show your working.

$4x^2 - x + 9 = 0$
provided possible answers:
$\frac{1}{4} +- \frac{i\sqrt143}{4}$ or $\frac{1}{8} +- \frac{i\sqrt143}{8}$

Mr F says: Show your working.

I know its kind of alot but I absolutely cannot figure this out! Help is very appreciated

..

**Drew_445** · November 11th 2008, 03:06 PM

Use the Quadratic Formula to solve the equation.

Mr F says: Show your working.

$x = \frac{5 +- \sqrt25-4(-2)(3)}{-4}$

$\frac{5 +- \sqrt1}{-4}$
So the solution I got was +-5 over 4, which doesn't answer any of the possible answers.

So what do i do?

**mr fantastic** · November 11th 2008, 03:29 PM

Originally Posted by Drew_445

Use the Quadratic Formula to solve the equation.

Mr F says: Show your working.

$x = \frac{5 +- \sqrt25-4(-2)(3)}{-4}$ Mr F says: Correct.

$\frac{5 +- \sqrt1}{-4}$ Mr F says: Wrong.

So the solution I got was +-5 over 4, which doesn't answer any of the possible answers.

So what do i do?

$x = \frac{5 \pm \sqrt{25 - 4(-2)(3)}}{-4} = \frac{5 \pm \sqrt{25 {\color{red}+} 24}}{-4} = \frac{5 \pm 7}{-4} = -3, ~ \frac{3}{4}$ .

**Drew_445** · November 11th 2008, 03:53 PM

oh wow such a simple solution

thanks very much

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