(x+2)(x-3)=(x-3)^2

- anonymous

(x+2)(x-3)=(x-3)^2

- katieb

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- anonymous

what is the short answer for this equation

- anonymous

solving for X I got 3/11 or .27 in fraction

- anonymous

how do u deal with it ?

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## More answers

- anonymous

well (x+2)(x-3) would be x-x-6 and (x-3)^2 would be x^2-6x+9. So x-x-6 = x^2-6x+9

- UsukiDoll

part of the FOIL process is wrong.

- UsukiDoll

for (x+2)(x-3) we have (x)(x)+(x)(-3)+2(x)+(2)(-3) which is x^2-3x+2x-6=x^2-x-6

- UsukiDoll

similarly, (x-3)^2 means write (x-3) twice based on the exponent number, so we have
(x-3)(x-3) = x^2-3x-3x+9 = x^2-6x+9

- anonymous

\[(x+2)(x-3) = (x-3)^2\]
\[(x+2)(x-3) = x ^{2}-x -6\]
\[(x-3)^{2}= x ^{2}-6x +9\]

- UsukiDoll

^ yes exactly..
\[x^2-x-6=x^2-6x+9\]

- UsukiDoll

is this a solve for x problem? like if I shift everything to the right we will have
\[x^2-x-6-x^2+6x-9=0\]

- UsukiDoll

since the x^2 cancels out can you tell me what -x+6x and -6-9 is?

- anonymous

@UsukiDoll
Did you mean -x +6x = -6x + 9

- UsukiDoll

but I'm shifting everything to the left... all signs from the right change.

- UsukiDoll

\[(x+2)(x-3) = (x-3)^2 \]
\[x^2-x-6=x^2-6x+9\]
\[x^2-x-6-x^2+6x-9=0 \]
the question op is asking is vague... so I'm not sure if this is a solve for x problem or something else

- anonymous

It will be simpler to solve by keeping the x on the left and the numbers on the right

- Zarkon

if you are solving for \(x\)
\[(x+2)(x-3)=(x-3)^2\]
\[(x+2)(x-3)-(x-3)^2=0\]
\[[(x+2)-(x-3)](x-3)=0\]
\[[x+2-x+3](x-3)=0\]
\[5(x-3)=0\]
so \(x=3\)

- UsukiDoll

yeah I was going to do that... since the x^2's cancel out I have \[5x-15 = 0 \rightarrow 5(x-3)=0\]

- Zarkon

or you do the following
\[(x+2)(x-3)=(x-3)^2\]
clearly x=3 is a solution...then cancel
\[x+2=x-3\]
cancel out the x's
so 2=-3 which is not possible...so x=3 is the only solution

- UsukiDoll

but again what is OP asking in here besides that question? It's just an equation. That can be anything out there.

- UsukiDoll

@LeibyStrauss I rather shift everything to the left and combine like terms... it's how I've done it when I learned the material. I don't want alternatives... it only creates drama between me and the person . just like in real life with the quadratic equation and my professor factored something out thinking that it will look simpler which was actually real hard.

- UsukiDoll

It didn't look simpler...it looked worse.

- UsukiDoll

wow @Zarkon that technique is actually much faster :O
because if I let x= 3 we have 3-3 on the left and right forcing the equation to look like 0=0

- anonymous

@UsukiDoll You taught me a new way how to solve it

- UsukiDoll

O_O! really? :D

- UsukiDoll

well...one of my professors always said that you can solve any math problem with different methods as long as it doesn't break the math rules.

- anonymous

Yes. I solved it as follows:
-x -6 = -6x + 9
Add 6x and 6 to both sides
5x = 15
divide both sides by 5
x = 15/5
We both got the same answer, but solving differently.

- anonymous

Let's solve your equation step-by-step.
(x+2)(x−3)=(x−3)2
Step 1: Simplify both sides of the equation.
x2−x−6=x2−6x+9
Step 2: Subtract x^2 from both sides.
x2−x−6−x2=x2−6x+9−x2
−x−6=−6x+9
Step 3: Add 6x to both sides.
−x−6+6x=−6x+9+6x
5x−6=9
Step 4: Add 6 to both sides.
5x−6+6=9+6
5x=15
Step 5: Divide both sides by 5.
5x5=155
x=3
Answer:
x=3

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