I've been working on this problem for forever and can't get it. 5x^2=-45x

I have to write it in standard form first i think..

Standard form? Like this: ax² + bx + c = 0 ?

the steps are 1. Write in standard form 2. Factor quadratic expression 3. set each factor to 0 and solve

Well, it can't get much more concise than that... to write it in standard, just put all the terms on one side, and 0 on the other.

exactly...so use the equation @terenzreignz has given....turn your equation into something like that

5x^2+45=0

One does not simply get rid of an x...

where did your x go from 45x?

Be very careful with manipulating equations. Human error is one of your worst enemies in this (noble) endeavour :D

sorry, I meant 5x^2+45x=0

There you go... now factor it :)

alright there we go.....now how would we factor it?

5x(x+45)=0?

And while you're thinking about it, an intermission, for any real numbers a and b if ab = 0 then a = 0 or b = 0 in other words, if the product of two numbers is zero, at least one of them must have been zero.

you factored out a 5x...but why not simplify the 45 x as well?

And uhh, your factored form, if we redistribute that term, we get 5x² + 135... try again :)

5x(x+5)=0?

how many times does 5 go into 45?

9 so, 5x(x+9)=0

there you go.....so now what?

Much better :) Now you have a product of two expressions equal to zero... sound familiar? Scroll up ^

5x=0 and x+9=0

Very good :) This gives you two solutions. What are they?

exactly....what are your zeros?

x=0 and x=-9

there ya go :)

Nicely done. Take this medal ;)

Thanks guys! I really appreciate it.

@terenzreignz sorry for the newbie question....but how do we give medals? is that what the "best response" is?

That's right :)

ahh okay thank you @terenzreignz

if the question had not been meant to teach a method ... i would have gone this route: 5x^2=-45x , x=0 makes this true 5x^2=-45x , for x ne 0, divide off an x 5x=-45 ; x=-9

I would have done the same, but they want me to practice the standard form

yeah :) and you did well

thank you!

what about this one? x^2=14x-49

I followed the same steps and got (x-7)(x-7) which would be x=7 and x=7 is this right? @terenzreignz @johnweldon1993 @amistre64

Rightly so. It has only one root (zero).

lol @terenzreignz beat me too it :)

This isn't really necessary, but it is possible to have a quadratic function with only one root. If written in this form... ax² + bx + c It WILL have only one root if \[\huge \sqrt{b^2-4ac}=0\]

Actually, that radical was unnecessary :D \[\huge b^2-4ac = 0\]...derp

yeahhh i was just going to say that lol

I beat you to it again... u mad, bro? :D

lmao really!? haha

x^2+18=-9x My answer: x=-6 and x=-3

p.s. Thanks for sticking with me through this. I'm not the best at math, but I try.

@terenzreignz @johnweldon1993

You're right. You're trying... and you're succeeding... more than you can say about others who try :D

Isn't one answer supposed to equal 0? I don't get 0 for either when I check my answer.

well lets try it out...(sorry did a midterm and had to switch classes) so we have x² + 18 = -9x add 9 x to both sides and we have... x² + 9x + 18 so what are the factors of 18? 18 and 1 9 and 2 6 and 3 which of those sets add to 9? looks like 6 and 3 right? so we have (x + 6)(x + 3) the zeros are indeed -6 and -3 lets plug them in (-6)² + 18 = -9(-6) 36 + 18 = 54 54 = 54 ...yes...that one checks out other one (-3)² + 18 = -9(-3) 9 + 18 = 27 27 = 27...indeed it does...so that is correct just recheck your math :)

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