Question

In solving the equation (x + 4)(x – 7) = -18, Eric stated that the solution would be
x + 4 = -18 => x = -22
or
(x – 7) = -18 => x = -11
However, at least one of these solutions fails to work when substituted back into the original equation. Why is that? Please help Eric to understand better; solve the problem yourself, and explain your reasoning

Answer 1

@JakeV8

Answer 2

I wouldn't go about it that way. Instead, expand the left hand side of the equation:
\[(x+4)(x-7) = x^2-3x-28\]
Now add 18 to both sides of the original equation:
\[x^2-3x-10 = 0\]
Now factor this to get an equation that is equal to zero. Then, the proper values for x will be clearly visible.

Answer 3

@LukeBlueFive :) Good thing you wouldn't go about it that way... the "Eric" in the problem was wrong :)

You can't solve a problem like his initial problem by considering only one parenthesis at a time to find the x values of the solution.

That works, but you have to have "zero" on the right side of the equation... he had "-18"

Answer 4

@JakeV8 explain lol

Answer 5

So @jdorta1, the way @LukeBlueFive shows is exactly right. What Eric should have done was first multiply out the stuff on the left, then get the "-18" over on the left also to leave just zero on the right.

Then, you can factor and solve for "x".

Answer 6

@JakeV8 ok so ya..how? lol

Answer 7

Does that make sense? what I said in words is precisely what @LukeBlueFive showed above.

Answer 8

Do you need help factoring the equation?

Answer 9

ugh math is crazy!

Answer 10

wow... lots of lag in the screen updates here :(

@jdorta1, how about you let @LukeBlueFive walk you through the steps?

Answer 11

@JakeV8 ok

Answer 12

so i now have (x+4)(x-7)=x^2-3x-28

Answer 13

Okay, to factor the equation we have:
\[x^2 - 3x - 10\]
Since we want to factor this into an equation of the form:
\[(x+a)(x-b)\]
We need the factors of 10, which are 2 & 5. Exactly one of them must be a negative. Which one should we choose, though? Well, the one where their sum will equal -3, so that we will get our middle term -3x.

Answer 14

gotta back up..how did you get a -10?

Answer 15

did i lose you guys?

Answer 16

(x + 4)(x – 7) = -18 <<<<---- Original
Now multiply out the left side:
x^2 + 4x -7x - 28 = -18
Simplify left side (combine x terms):
x^2 - 3x - 28 = -18
Now add 18 to both sides to get rid of the -18 on the right side:
x^2 - 3x - 28 + 18 = -18 + 18
Simplify:
x^2 - 3x -10 = 0

Now you can factor :)

Answer 17

@JakeV8 ok so how do you factor?

Answer 18

Rather than re-type it, can you review what @LukeBlueFive a few responses ago? that was a good explanation of how to factor.

Answer 19

@JakeV8 so it would be (x+2)(x-5)

Answer 20

yes... if you multiply that out, it matches what you had...

(x+2)(x-5) = x^2 +2x - 5x - 10 which is just x^2 -3x -10... so that's correctly factored.

Answer 21

ok so is that the answer then?

Answer 22

Now you just need to find what values of x make (x+2)(x-5) = 0.

Answer 23

almost :)

When you have a factored form on the left and " = 0" on the right, you are close... you just need the solution values for x.

(x+2)(x-5) = 0

If either (x+2) or (x-5) is = 0, then the stuff on the left multiplies to 0, making the equation true.

(x+2) = 0 when x = -2
and
(x-5) = 0 when x = 5

So those are the solutions.

Answer 24

ooooooooh gotcha!

Answer 25

yay thank you guys so much!

Answer 26

So, take one quick look at the original problem...

Answer 27

ok

Answer 28

The reason Eric's approach was wrong was that he did that same thing I just did, but he did it with -18 on the other side of the equals sign.

Answer 29

It has to be zero on the other side.

Answer 30

ok

Answer 31

@jdorta1 You're very welcome! Glad I could help. BTW, nice assist @JakeV8!

Answer 32

so the answer is y=-2 and x=5

Answer 33

So @LukeBlueFive showed the way to get zero on the right side... multiply everything out on the left, then move the -18 over, then refactor it all.

Once it's factored on the left, and all = zero on the right, then you can do what I just showed you to find the solutions for x.

Answer 34

sweet! so who wants the medal thingy lol you guys were both a great help

Answer 35

Give it to @LukeBlueFive :)

Answer 36

Jake should get it. I had to bail half-way through.

Answer 37

ok =)

Answer 38

too late haha @LukeBlueFive got it

Answer 39

lol, well, thank you! ^_^

Answer 40

=) very welcome.

Answer 41

hope to have you guys help me again soon =) @JakeV8 and @LukeBlueFive

Answer 42

Glad to help... :)

Answer 43

=)

Answer 44

Hope we're available when you need help. =)