Question

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

Answer 1

what is the short answer for this equation

Answer 2

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

Answer 3

how do u deal with it ?

Answer 4

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

Answer 5

part of the FOIL process is wrong.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

^ yes exactly..

\[x^2-x-6=x^2-6x+9\]

Answer 10

is this a solve for x problem? like if I shift everything to the right we will have

\[x^2-x-6-x^2+6x-9=0\]

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

\[(x+2)(x-3) = (x-3)^2 \]
\[x^2-x-6=x^2-6x+9\]
\[x^2-x-6-x^2+6x-9=0 \]

the question op is asking is vague... so I'm not sure if this is a solve for x problem or something else

Answer 15

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

Answer 16

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 \(x=3\)

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

@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.

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

O_O! really? :D

Answer 25

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.

Answer 26

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.

Answer 27

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