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## anonymous one year ago (x+2)(x-3)=(x-3)^2

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1. anonymous

what is the short answer for this equation

2. anonymous

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

3. anonymous

how do u deal with it ?

4. 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

5. UsukiDoll

part of the FOIL process is wrong.

6. 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

7. 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

8. anonymous

$(x+2)(x-3) = (x-3)^2$ $(x+2)(x-3) = x ^{2}-x -6$ $(x-3)^{2}= x ^{2}-6x +9$

9. UsukiDoll

^ yes exactly.. $x^2-x-6=x^2-6x+9$

10. UsukiDoll

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

11. UsukiDoll

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

12. anonymous

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

13. UsukiDoll

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

14. 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

15. anonymous

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

16. 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 $$x=3$$

17. UsukiDoll

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

18. 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

19. UsukiDoll

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

20. 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.

21. UsukiDoll

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

22. 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

23. anonymous

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

24. UsukiDoll

O_O! really? :D

25. 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.

26. 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.

27. 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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