How do I solve for X?

X^2+7x=18

$$x^2+7x-18=0 \qquad (x-2)(x+9)=0\qquad x_1=2 \quad x_2=-9$$

You start by taking it all to one side 

$$\\X^2+7x=18\\
X^2+7x-18=0$$

now I would like either the question asker or Tenacious to factorize the left hand side for mw please.

Hint1:  You need 2 numbers that mult to -18 and add to +7 

Hint2:  What will the signs of these numbers be?

is no one going to answer this?

I've started answering it.  Now I have asked you to try the next bit.  

x²+7x=18

x²+7x-18=0

(x-2)(x+9)=0  So x=+2 or x=-9   Edited because I had the + and - the wrong way round originally, now correct. Thanks to Melody for pointing that out.

x²+9x-2x-18

x²+7x-18

By the time I saw this there were other answers so too late but I purposefully avoided looking at them, no point in cheating. I only saw the first two posts, the question and Melody's hints. I doubt I would have done this so easily without those hints Melody, so thanks. I think this is the first time I have encountered a quadratic trinomial in a practical ish situation. Up till now I have only encountered them as isolated exercises. It is useful to see how they can actually be used and be of benefit.

hi Tenacious,

Your third line is missing =0

The conclusion you have drawn is incorrect.

Heureka's answer is correct.  See if you can work out the logic of why.

We may need to discuss it.

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Your fourth and fifth lines are just checking that the factorising is correct so if you want to include it at all you should insert the words "checking factorization ".  
Or something like that. :)   

You can check the finished answer - that would be better anyway.

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I'll let you think about it a bit and then I will give you the full worked answer with explanation if you want me to.     

Ah yes, I was in a rush as was just on my way out and the first time I tried to publish it just went back to the question so had to retype it all. Must have retyped it incorrectly as it obviously doesn't work in that format with + and - in wrong order. Should have read;

x²+7x=18

x²+7x-18=0

(x-2)(x+9)=0  So x=+2 or x=-9

I always expand the brackets after factorising to double check and usually substitute the algebraic terms for my answers to double check as well. I rarely trust myself enough to just put the answers without double checking them. 

Hi Tenacious,

That is good :)

So you know how to substitute to check if  2 and -9 are correct ?

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Would you like another?

$$3x^2+36=24x$$

It is the same except you need to factor out the common factor before you factor the rest of the trinomial.

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Also, how do you put the hyperlinks in?   I do not know how to do that.   

Thanks  

Yeh, substituting. Before I knew enough to be able to do this sort of thing as close to properly as I do, I used to just play around with substitution as it was the only way I could work out how to do it with common sense alone. Sometimes it could take a while. For years I developed my own common sense coping strategies for maths as I didn't know the correct methods. I very much doubt I came up with anything new but I did come up with them myself, they were never very quick though. You may see that I do some things a little bit unconventionally.

3x²+36=24x    Subtract 24x from both sides;

3x²-24x+36=0    Not sure if I needed to put the "-24x" in this position but it normally sits here as B in the formula so thought it best? Then I factorised;

3(x²-8x+12)=0    Then I divided both sides by 3 which seemed correct for the left hand side but seemed like a bit of a cheat for the right hand side as $${\frac{{\mathtt{0}}}{{\mathtt{3}}}}$$ stays as 0, but it seems to work;

x²-8x+12=0   Then I factorised again;

(x-6)(x-2)=0   So x=6 or x=2

I have expanded the brackets and substituted the numbers and all seems to check ok. This really does help to make sense of factorising, it gives it a purpose.

I'm still never satisfied though to come up with an answer of it could be this or it could be that, I've always been used to there only really being one correct answer in maths, different ways of getting there but only one correct answer.

Forgot about the hyperlink.

Just click the chain symbol in the tool bar (though you could also say it looks like a head on view of a pair of glasses at 45 degrees), just under the background colour selector. I'm sure you know or can guess the rest but just in case, the "URL" is the address from the address bar and "Text to display" is where you put "Click here" or whatever text you want to appear as your hyperlink in your post. Sorry if that was teaching you to suck eggs (not sure if that's a term common in your neck of the woods)(oops, there's another one).

Thanks Tenacious.  I shall have to give that hyperlink a go :)

your factorising was done beautifully.  Yes the trinomial must be in order.  $$x^2$$    first then $$x$$  then the constant.  

Here is a slightly different type of factorising.  they are not hard - see if you can work it out - maybe you already know.

HINT:    factorise  5xy-2y     Can you see the similarity?    

1)  5x(x+3)-2(x+3)  

2)   -7x(2x-1)-(2x-1)

I am going to test that hyperlink here

My country

lets see if this works

It does   

Thanks Tenacious - I am really pleased 

The hyperlink is a lot more elegant than having to have the whole web address in your post, glad it worked for you.

1) 5x(x+3)-2(x+3)

= (x+3)(5x-2)   I guess this is my answer. Not that familiar with this, seemed a bit too simple so worked it the other way to check as usual. Expanded the brackets;

= 5x²-2x+15x-6   Then combined like terms;

= 5x²+13x-6      Then tried to factorise this, looking for something that would multiply to make -30 and add to make 13;

= 5x²-2x+15x-6   Continued factorising and got back to your original question so should be good.

2) -7x(2x-1)-(2x-1)

= (2x-1)(-7x-1)   I think this is my answer. Threw me a bit only having the - in front of the second bracket with no number, but went through the same process and it seems to work ok. Expanded;

= -14x²-2x+7x+1   Combined like terms;

= -14x²+5x+1   Factorise, look for something that multiplies to -14 and adds to 5.

= -14x²+7x-2x+1   Then back to where we started again.

I am quite slow on these. I think I used to just struggle playing around with numbers in brackets until it worked but this is a more deliberate method. 

I am a bit thrown by your hint "HINT:    factorise  5xy-2y     Can you see the similarity?" Was this to relate to question 1?

HINT:    factorise  5xy-2y

well here you have   5x lots of y minus  2 lots of y  so you must end up with (5x-2) lots of y

5xy-2y = (5x-2)y   or    y(5x-2)

consider 2x+3x = 2lots of x + 3 lots of x = (2+3)lots of x = 5x   [you knew the answer was 5x in the first place]

so

6(x+3)+2(x+3) = (6+2)(x+3)=8(x+3)

7(y-4)+y(y-4)=(7+y)(y-4)

These facctorizations questions are very easy.  Do you understand what I am trying to show you?