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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    what is the short answer for this equation

  2. anonymous
    • one year ago
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    solving for X I got 3/11 or .27 in fraction

  3. anonymous
    • one year ago
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    how do u deal with it ?

  4. anonymous
    • one year ago
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    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
    • one year ago
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    part of the FOIL process is wrong.

  6. UsukiDoll
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \[(x+2)(x-3) = (x-3)^2\] \[(x+2)(x-3) = x ^{2}-x -6\] \[(x-3)^{2}= x ^{2}-6x +9\]

  9. UsukiDoll
    • one year ago
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    ^ yes exactly.. \[x^2-x-6=x^2-6x+9\]

  10. UsukiDoll
    • one year ago
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    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
    • one year ago
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    since the x^2 cancels out can you tell me what -x+6x and -6-9 is?

  12. anonymous
    • one year ago
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    @UsukiDoll Did you mean -x +6x = -6x + 9

  13. UsukiDoll
    • one year ago
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    but I'm shifting everything to the left... all signs from the right change.

  14. UsukiDoll
    • one year ago
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    \[(x+2)(x-3) = (x-3)^2 \] \[x^2-x-6=x^2-6x+9\] \[x^2-x-6-x^2+6x-9=0 \] 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
    • one year ago
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    It will be simpler to solve by keeping the x on the left and the numbers on the right

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

  20. UsukiDoll
    • one year ago
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    @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
    • one year ago
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    It didn't look simpler...it looked worse.

  22. UsukiDoll
    • one year ago
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    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
    • one year ago
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    @UsukiDoll You taught me a new way how to solve it

  24. UsukiDoll
    • one year ago
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    O_O! really? :D

  25. UsukiDoll
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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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