How might one go about doing this problem? Solve for x: x^2 + 24x + 90 = 0

- poopsiedoodle

How might one go about doing this problem? Solve for x: x^2 + 24x + 90 = 0

- Stacey Warren - Expert brainly.com

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- anonymous

Factor first :)

- anonymous

What factors of 90, when added together, give you 24?

- poopsiedoodle

Gimme a second.

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## More answers

- anonymous

Take your time (:

- poopsiedoodle

I don't think any do. 18 and 5 is the closest you can get to 24, which is 23.

- anonymous

Hmm, hold on a sec

- poopsiedoodle

I will do that.

- anonymous

Well, since it isn't factorable, we have to use the quadratic formula. Have you learned it?

- poopsiedoodle

Hardly. I've just started on it.

- anonymous

Well, this may take some explaining to do, so hold on for a bit haha.

- anonymous

x^2 + 24x + 90 = 0 Let's break down what this problem means. By plugging in a certain number(s) for x, you get zero. It's pretty much like plotting a graph of x^2 + 24x + 90 and looking for where the graph touches zero. Did I lose you yet?

- anonymous

|dw:1358120873885:dw|

- poopsiedoodle

Okay. I get it so far.

- anonymous

Alright. So since we have an equation that isn't factorable, we find the zeroes by plugging it into the quadratic formula.

- anonymous

This is the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac} }{ 2a }\] Now, a, b, and c are the numbers in your equation -- \[a^2+bx+c\] In this case: \[x^2 + 24x + 90\] so, a=1, b= 24, and c=90.

- poopsiedoodle

Ok, lemme try it now.

- anonymous

Remember, the ± symbol indicates that there are two expressions! (ex. 1 ± 3 is 1 + 3 AND 1 - 3)

- poopsiedoodle

\[\Large0= \frac{ -24 \pm \sqrt{24^{2} -4(90)} }{ 2 }\]

- anonymous

Good so far!

- anonymous

except, it's x =

- poopsiedoodle

Well, since we're already using x in the original problem, how about Y?

- anonymous

But we're trying to solve for x, haha

- poopsiedoodle

eh, whatever. By the way, I'm having a slight problem with \[\sqrt{4(90)}\]

- anonymous

That's just \[\sqrt{4 • 90} = \sqrt{360}\]

- poopsiedoodle

Ah. \[\Large x= \frac{ -24 \pm 24\sqrt{360} }{ 2 }\]

- anonymous

No, remember 24^2 is under the radical (Don't take it out!) \[\sqrt{24^2 - 360}\]

- poopsiedoodle

\[\Large x= \frac{ -24 \pm \sqrt{ 24^{2} (-360)} }{ 2 }\]*

- poopsiedoodle

But, since √24^2 = 24, what's wrong with taking it out?

- anonymous

Because we can't have a negative number under a square root, haha

- poopsiedoodle

Ah.

- anonymous

But anyways, after plugging all of those numbers in, what do you geT?

- poopsiedoodle

So, \[\Large x= { -12 \pm (\sqrt{ 24^{2} (-360)} /2)}\] ? I'm confused now e_o

- anonymous

Well, the /2 should be on the bottom of the whole thing, but besides that, Let's simplify the square root first. What is 24^2 - 360?

- poopsiedoodle

216.

- poopsiedoodle

and the root of that is 14.giganticdecimal

- anonymous

Well, let's simplify it haha sqrt of 216 can be reduced into what? (hint: 36 x 6 = 216)

- poopsiedoodle

oh gee, this is a tought one... 2... and... erm... GIMME A MINUTE

- poopsiedoodle

tough*

- poopsiedoodle

OOH! I KNOW! 36 AND 6. BOOM. Don't you wish you had thought of that?

- anonymous

Hahahaha. You are correct x)

- anonymous

So... now we have \[x = \frac{ -24 \pm 6\sqrt{6} }{ 2 }\] which breaks down to: \[x = \frac{ -24 + 6\sqrt{6} }{ 2 }\] \[x = \frac{ -24 - 6\sqrt{6} }{ 2 }\] I assume you can solve from there :)

- poopsiedoodle

\[\Huge x = { -12 - 3\sqrt{3} }\] ?

- anonymous

Annnnnnnnnd?

- poopsiedoodle

I was wondering if I was right, which I guess I am. So, is it \(\Huge x=-12 -3?\)

- anonymous

Wait, no don't take out the radical 3 lol You need to solve for the other part: \[x = \frac{ -24 + 6\sqrt{6} }{ 2 }\] Your first solution was correct, btw!

- poopsiedoodle

Oh. x=-12 + 3√3. And now what do I do?

- anonymous

Well, those are your solutions, haha x=-12 + 3√3 and x=-12 - 3√3 Yay!

- poopsiedoodle

:D

- anonymous

:D indeed! If you need more help, just pm me (:

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