anonymous
  • anonymous
(x+2)(x-3)=(x-3)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
what is the short answer for this equation
anonymous
  • anonymous
solving for X I got 3/11 or .27 in fraction
anonymous
  • anonymous
how do u deal with it ?

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More answers

anonymous
  • 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
  • UsukiDoll
part of the FOIL process is wrong.
UsukiDoll
  • 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
  • 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
  • anonymous
\[(x+2)(x-3) = (x-3)^2\] \[(x+2)(x-3) = x ^{2}-x -6\] \[(x-3)^{2}= x ^{2}-6x +9\]
UsukiDoll
  • UsukiDoll
^ yes exactly.. \[x^2-x-6=x^2-6x+9\]
UsukiDoll
  • 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
  • UsukiDoll
since the x^2 cancels out can you tell me what -x+6x and -6-9 is?
anonymous
  • anonymous
@UsukiDoll Did you mean -x +6x = -6x + 9
UsukiDoll
  • UsukiDoll
but I'm shifting everything to the left... all signs from the right change.
UsukiDoll
  • 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
  • anonymous
It will be simpler to solve by keeping the x on the left and the numbers on the right
Zarkon
  • 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
  • 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
  • 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
  • UsukiDoll
but again what is OP asking in here besides that question? It's just an equation. That can be anything out there.
UsukiDoll
  • 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
  • UsukiDoll
It didn't look simpler...it looked worse.
UsukiDoll
  • 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
  • anonymous
@UsukiDoll You taught me a new way how to solve it
UsukiDoll
  • UsukiDoll
O_O! really? :D
UsukiDoll
  • 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
  • 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
  • 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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