# Thread: Simple factorization/equation problems. Need help!

1. ## Simple factorization/equation problems. Need help!

(Posted this on mymathforum.com as well. This is a copy-paste)

I'm a Swedish student, reading course C in mathematics. I've read course A and B before, but it was more than five years ago, and I've forgot about everything (they predict that everyone knows course A and B).

I'm having problem with factorization and equations.

Equations
:
I can solve very easy equations like x^2-4x+3 and stuff like that. What I'm having trouble with are √-equations.
I can solve for example √(2x+5)=x+1 by doing this:
1. 2x+5 = (x+1)^2
2. 2x+5 = x^2+2x+1
3. x^2-4 = 0
4. x=2

But then, when I come to this problem, √(4t+1)=3-3t, I can't solve it:
1. 4t+1 = (3-3t)^2
2. 4t+1 = 9-18t+9t^2
3. 9t^2-22t+8 = 0
4. What here? The answer is (4/9). I know that I somehow should divide everything with 9. Should I first calculate everything and the divide by 9?

Two more problems: 3√(x+13) = x+9 & √(x+1)-1=x. How do I solve this?

Factorizations:

In short: How do I get the answer 7(x-1)(x+2/7) from 7x^2-5x-2? Are there any steps/formulas to get to this answer? How do I know that 7x^2-5x-2 should be written in the form a(b-c)(b+d), and not something like a(a-b-c) etc. When I know that it should be written like a(b-c)(b+d), how do I calculate what should be in "c" and in "d"?

I hope I've explained my problems well. As said, I haven't done any mathematics in over five years, and the course expect me to know course A and B already, and I feel totally lost.

2. ## Response

1. 4t+1 = (3-3t)^2
2. 4t+1 = 9-18t+9t^2
3. 9t^2-22t+8 = 0

For this one use the quadratic formula:

$
x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}
$

So, it would be:
$
x = \frac{-(-22)\pm\sqrt{(-22)^2-4*9*8}}{2*9}
$

$
x = \frac{22\pm\sqrt{484-288}}{18}
$

$
x = \frac{22\pm\sqrt{196}}{18}
$

$
sqrt{196} = 14
$

t = (22 + 14) / 18 or t = (22 - 14) / 18
t = 36/18 or t = 8/18
t = 2 or t = 4/9

OR, you could divide the original equation through by 9 to get:
t^2 - (22/9)t + 8/9 = 0

Then factor it as:
(t-2)(t-4/9) = 0
but, these factors might not have been so obvious. When in doubt, use the quadratic formula to find the values

3. ## Response

3√(x+13) = x+9
Start by squaring both sides to get:
9(x+13) = (x+9)^2
9x + 117 = x^2 + 18x + 81
0 = x^2 + 18x + 81 - 9x - 117
x^2 + 9x -36 = 0

This one can be factored of the form: (x-a)(x+b)
Try some of the factors of 36 to be substituted into a,b, namely 12 and 3

(x + 12)(x - 3) = x^2 - 3x + 12x -36 = x^2 + 9x - 36 = 0
So, you would have:
x = -12 or x = 3

When in doubt you could always use the quadratic formula to get the solutions, like I did in the last problem.

For some reason, when I substitute x = -12 back into the original equation, I get 3 = -3. I've checked and can't find the error. Maybe you could go over this one with your teacher or a classmate.

4. ## Response

√(x+1)-1=x

For this one, start by adding 1 to both sides:
√(x+1) = x+1 Now, square both sides to get:
x+1 = (x+1)^2
x+1 = x^2 + 2x + 1
0 = x^2 + 2x + 1 - x - 1
x^2 + x = 0

Factor out an x:
x(x + 1) = 0

Solve for x:
x = 0 or x = -1

I can solve very easy equations like x^2-4x+3 and stuff like that. What I'm having trouble with are √-equations.
I can solve for example √(2x+5)=x+1 by doing this:
1. 2x+5 = (x+1)^2
2. 2x+5 = x^2+2x+1
3. x^2-4 = 0
4. x=2

When you get it down to step 3, you need to factor it as:
(x-2)(x+2) = 0

x = 2 or x = -2.

Substitute x = -2 back into the original equation in step 1. and you will see that it checks out.

6. ## Last part

How do I get the answer 7(x-1)(x+2/7) from 7x^2-5x-2? Are there any steps/formulas to get to this answer? How do I know that 7x^2-5x-2 should be written in the form a(b-c)(b+d), and not something like a(a-b-c) etc. When I know that it should be written like a(b-c)(b+d), how do I calculate what should be in "c" and in "d"?

Well, you would start by factoring out a 7 from the equation:
7(x^2 - (5/7)x - 2/7)
Then you would factor it as such:
7(x-1)(x+2/7)

This sometimes takes guess and check work. If you wanted to, you could use the quadratic formula, find the x values, and then write it of this form.
I hope all of this helps.

7. Wow, thank you ajj86 for your help! You have really helped me out