# Thread: Another qaudratic equation

1. ## Another qaudratic equation

I'm finding it difficult to solve this question
It's solving quadratic eqaution by completing the square

x^2+12x+3=0

This is what I have done so far ...

x^2+12x+3=0
x^2 +12x=-3
(x+12)^2-144=-3
(x+12)^2=141
x+12=√141

Then, I don't know what to do

2. Originally Posted by Mist
I'm finding it difficult to solve this question
It's solving quadratic eqaution by completing the square

x^2+12x+3=0

This is what I have done so far ...

x^2+12x+3=0
x^2 +12x=-3 .. its correct till here .
(x+12)^2-144=-3 it should be halved
(x+12)^2=141
x+12=√141

Then, I don't know what to do
HI

$\displaystyle x^2+12x+3=0$

$\displaystyle x^2 +12x=-3$

$\displaystyle (x+6)^2=-3+6^2$

$\displaystyle (x+6)^2=33$

$\displaystyle x=\pm\sqrt{33}-6$

3. Originally Posted by Mist
I'm finding it difficult to solve this question
It's solving quadratic eqaution by completing the square

x^2+12x+3=0

This is what I have done so far ...

x^2+12x+3=0
x^2 +12x=-3
(x+12)^2-144=-3
(x+12)^2=141
x+12=√141

Then, I don't know what to do
You have the right idea. However, it should be $\displaystyle (x+6)^2$, not $\displaystyle (x+12)^2$, you always take half of the coefficient of $\displaystyle x$.

Also, when you square root your equation, you end up with two terms on the right hand side. For example $\displaystyle \sqrt{4}$ would give you 4 and -4.

You seem to be struggling with some of the basic mathematical operations, I suggest that you take some time to read through your notes/books on the subject.

Hope this helps

4. Have I solved this question correctly?

x^2-3x-5=0
(x-1.5)^2-2.25=5
x-1.5=√7.25
x=1.5√7.25

The answer seems wrong

5. Originally Posted by Mist
Have I solved this question correctly?

x^2-3x-5=0
(x-1.5)^2-2.25=5
x-1.5=√7.25
x=1.5√7.25
The answer seems wrong
No.

You started off right, however you went wrong from the highlighted point above.

Because you are only considering the positive square root you are missing out a solution!

I have $\displaystyle x = 1.5 \pm\sqrt{7.25}$

You are not far off, just keep practising, and take your time, dont rush things when you are unsure about them.

Hope this helps

6. Originally Posted by Mist
Have I solved this question correctly?

x^2-3x-5=0
(x-1.5)^2-2.25=5
x-1.5=√7.25 -- you missed the +/- in front of the square root .
x=1.5√7.25 -- its + 1.5

The answer seems wrong
The rest is correct .

7. Originally Posted by Mist
x=1.5√7.25
Also here, do you mean $\displaystyle 1.5 + \sqrt{7.25}$ or $\displaystyle 1.5 \times \sqrt{7.25}$ ?

8. ## Stuck on another question

How would I solve this?

2x^2-7=4x

This is what I have done:
2x^2-7=4x
2x^2-4x-7=0
2(x-2)^2+4=7
2(x-2)^2=11

I don't know what to do after that

9. Originally Posted by Mist
How would I solve this?

2x^2-7=4x

This is what I have done:
2x^2-7=4x
2x^2-4x-7=0
2(x-2)^2+4=7
2(x-2)^2=11

I don't know what to do after that
If there is a coefficient other that 1 in front of the $\displaystyle x^2$ term you need to remove it before completing the square.

In the case of your equation, before you start rearranging the equation divide everything by 2, then proceed as normal.

10. ## What I get

x^2+12x+3=0

(x+6)^2-33=0

(x+6)^2=33

x+6= rt33

x= 6+- rt33

p.s is there a way to get all the maths signs? I don't really wanna have to write rt lol.

11. you forgot to change the 6 to a negative number when you moved it to the other side of the equal sign

$\displaystyle x = -6 \pm\sqrt{33}$

In addition, I need to learn how to properly format text

12. Originally Posted by RRH
you forgot to change the 6 to a negative number when you moved it to the other side of the equal sign

$\displaystyle x = -6 \pm\sqrt{33}$

In addition, I need to learn how to properly format text
Oh yehh darn , shouldn't have made that mistake.

13. Originally Posted by alibond07
...

p.s is there a way to get all the maths signs? I don't really wanna have to write rt lol.
Go to the forum LaTex Help and have a look here: http://www.mathhelpforum.com/math-he...-tutorial.html