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Mathematics 14 Online
OpenStudy (anonymous):

What is the solution to the equation? HELLLP!!!

OpenStudy (anonymous):

\[\sqrt{2X+13} - 5 = X\]

OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (misty1212):

HI!!

OpenStudy (misty1212):

1) add \(5\)

OpenStudy (anonymous):

Hi! Can you please help?

OpenStudy (misty1212):

2) square

OpenStudy (anonymous):

to both sides?

OpenStudy (misty1212):

go ahead and do those, i will check yes of course what you do to one side, you do to the other

OpenStudy (anonymous):

\[\sqrt{2x+13}-5=x\]\[\sqrt{2x+13}=5x\]

OpenStudy (anonymous):

right?

OpenStudy (misty1212):

oh no dear, don't multiply \(5x\) means \(5\times x\) you want to ADD

OpenStudy (anonymous):

\[\sqrt{2x+13}=x+5\]

OpenStudy (anonymous):

? like that?

OpenStudy (misty1212):

yes

OpenStudy (anonymous):

now what? I dont understand what you mean by square? And thanks

OpenStudy (misty1212):

much better now square both sides so you get rid of the radical on the left

OpenStudy (misty1212):

square means to multiply it by itself you need to get rid of the square root, so you do that by squaring it

OpenStudy (anonymous):

So\[2x ^{2}+13^{2}=x^{2}+5^{2}\] \[4x+169=x^{2}+25\]

OpenStudy (anonymous):

?

OpenStudy (anonymous):

or would the 4x be \[4x^{2}\]

OpenStudy (misty1212):

ok lets go back up slowly q

OpenStudy (misty1212):

we got \[\sqrt{2x+13}=x+5\] when you square the left, you just get \(2x+13\)

OpenStudy (misty1212):

in other words, the radical goes bye bye that is the whole point

OpenStudy (misty1212):

so we get \[2x+13=(x+5)^2\] as the next step

OpenStudy (anonymous):

Oh okay! I see. then \[x ^{2}+25\]

OpenStudy (anonymous):

or no?

OpenStudy (misty1212):

oooh no don't make that mistake or your teacher will think you are a moron you don't want that !

OpenStudy (anonymous):

lol.. okay.. then what do i do?

OpenStudy (misty1212):

\[(x+5)^2=(x+5)(x+5)=x^2+5x+5x+25=x^2+10x+25\]

OpenStudy (anonymous):

wait what?

OpenStudy (misty1212):

square means multiply by itself

OpenStudy (misty1212):

some teachers like to call that "foil" maybe you heard it that way

OpenStudy (anonymous):

yeah, and x * x is x^2? and 5*5 is 25? i thought...

OpenStudy (misty1212):

yea i know what you thought you thought you could square each piece separately

OpenStudy (misty1212):

and yes, there is an \(x^2\) in the answer, as well as a \(25\) but it is not \[(x+5)^2=x^2+25\] like i said DO NOT MAKE THAT MISTAKE it is \((x+5)^2=x^2+10x+25\)

OpenStudy (anonymous):

okay. now what?

OpenStudy (misty1212):

ok now it has been so long we start at the top and go quicker \[\sqrt{2x+13} - 5 = x\] add 5\[\sqrt{2x+13}=x+5\] square (correctly)\[2x+13=x^2+10x+25\] now you have a quadratic equation to solve

OpenStudy (misty1212):

subtract \(2x\) and \(13\) to set it equal to zero, get \[x^2+8x+12=0\] and now solve for \(x\) by factoring if you can

OpenStudy (anonymous):

\[8x^{2}=-12?\]

OpenStudy (anonymous):

or no?

OpenStudy (misty1212):

no

OpenStudy (anonymous):

or \[x^{2} + 8x=-12\]

OpenStudy (anonymous):

oh no?

OpenStudy (misty1212):

do you know how to factor?

OpenStudy (anonymous):

No. :/

OpenStudy (misty1212):

hmm that is a problem because you really want to turn \[x^2+8x+12=0\] in to \[(x+2)(x+6)=0\]

OpenStudy (anonymous):

How did you get 2 and 6?

OpenStudy (misty1212):

have you spent any time solving quadratic equations ?

OpenStudy (misty1212):

i got them because \(2\times 6=12\) and \(2+6=8\)

OpenStudy (anonymous):

No, I missed 3 years of highschool because I had lynch syndrome(cancer), I was NEVER there to learn anything. I am pretty much teaching myself through the course of online schooling just trying to pass and get to college. and I see. So you find two numbers that can go into both of them?

OpenStudy (misty1212):

two numbers whose product is 12 and whose sum is 8 yes

OpenStudy (anonymous):

okay, i understand. so then do we do that foil thing you talked about?

OpenStudy (misty1212):

now that we have \[(x+2)(x+6)=0\] we solve \[x+2=0\] and get \(x=-2\) and also \(x+6=0\) to get \(x=-6\)

OpenStudy (misty1212):

then we go back to the original equation and try them out

OpenStudy (anonymous):

okay, I see. so we plug it in?

OpenStudy (anonymous):

\[\sqrt{2(-2)+13}-5=-6\]

OpenStudy (anonymous):

?

OpenStudy (misty1212):

yes \[\sqrt{2x+13}-5=x\] \[\sqrt{2\times (-2)+13}-5=-2\] true for faclse

OpenStudy (anonymous):

false? Shouldnt it be =-6?

OpenStudy (misty1212):

the right sides should be \(-2\) because it is x

OpenStudy (misty1212):

we will try \(-6\) second

OpenStudy (anonymous):

okay

OpenStudy (misty1212):

now we are trying \(x=-2\) which means everywhere

OpenStudy (anonymous):

okay I see. so \[\sqrt{2(-2)+13}-5=-2\]

OpenStudy (misty1212):

\[\sqrt{2\times (-2)+13}-5=-2\] works because you get \[\sqrt{-4+13}-5=\sqrt{9}-5=3-5=-2\checkmark\]

OpenStudy (anonymous):

Okay, so is that all we do?

OpenStudy (misty1212):

no, now we try \(x=-6\) to see if that works

OpenStudy (misty1212):

\[\sqrt{2(-6)+13}-5=-6?\]

OpenStudy (anonymous):

\[\sqrt{-12+13} -5=-6\]

OpenStudy (misty1212):

this one does not work we get \[\sqrt{1}-5=1-5=-4\neq -6\]

OpenStudy (anonymous):

so it would be -2?

OpenStudy (misty1212):

whew yes, after all that

OpenStudy (misty1212):

so lets see what we needed to do this a) a method b) knowing how to square correctly c) solving a quadratic equation d) knowing to check the answer

OpenStudy (anonymous):

So that is the final answer?

OpenStudy (misty1212):

yes final answer is -2

OpenStudy (anonymous):

Thank you soooo much.

OpenStudy (misty1212):

i would suggest that the other stuff is somewhat more important than this but no matter you do what you have to your welcome \(\color\magenta\heartsuit\)

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