# Thread: calculation to an equation?

1. ## calculation to an equation?

$\displaystyle 4*(5x+3)^2=9$

I am havning problems with the process to solve this Equation.
This is how far i have come:

$\displaystyle 4*(5x+3)^2=9$
$\displaystyle 4*(5x^2+30x+9)=9$ Using a rule here that i dont know the name of in english
$\displaystyle 20x^2+120x+36=9$
$\displaystyle 20x^2+120x=-27$
$\displaystyle 20x^2+x=-27/120$ now what? How do I get rid of the "^2" from 20x? or is this totaly wrong?

Fast answers are very much appreciated here!

Rickard Liljeros

2. Hello,

Originally Posted by liljeros
$\displaystyle 4*(5x+3)^2=9$

I am havning problems with the process to solve this Equation.
This is how far i have come:

$\displaystyle 4*(5x+3)^2=9$
$\displaystyle 4*(5x^2+30x+9)=9$ Using a rule here that i dont know the name of in english << SiMoon says : "expanding" or "developping" or the quadratics formula ?
$\displaystyle 20x^2+120x+36=9$
$\displaystyle 20x^2+120x=-27$
$\displaystyle 20x^2+x=-27/120$ now what? How do I get rid of the "^2" from 20x? or is this totaly wrong?

Fast answers are very much appreciated here!

Rickard Liljeros
Don't bother

$\displaystyle 4=2^2$

$\displaystyle \implies 4*(5x+3)^2=\left(2(5x+3)\right)^2=\boxed{(10x+6)^2 }$

Therefore :
$\displaystyle 4*(5x+3)^2=9 \Longleftrightarrow (10x+6)^2=9 \Longleftrightarrow (10x+6)^2-3^2=0$

Now, remember that $\displaystyle a^2-b^2=(a-b)(a+b)$

And it's all done

3. Thank you very much! That helped me alot, I need to learn to try out more then one way in the begining

4. Originally Posted by Moo
Hello,

Don't bother

$\displaystyle 4=2^2$

$\displaystyle \implies 4*(5x+3)^2=\left(2(5x+3)\right)^2=\boxed{(10x+6)^2 }$

Therefore :
$\displaystyle 4*(5x+3)^2=9 \Longleftrightarrow (10x+6)^2=9 \Longleftrightarrow (10x+6)^2-3^2=0$

Now, remember that $\displaystyle a^2-b^2=(a-b)(a+b)$

And it's all done
Now I fond one thing here i didnt understand, when u multiply the "4" in to the "()" you dont get 10 and 6? isnt 5*4=20 and 4*3=12

5. Originally Posted by liljeros
Now I fond one thing here i didnt understand, when u multiply the "4" in to the "()" you dont get 10 and 6? isnt 5*4=20 and 4*3=12
No, it's $\displaystyle 4*(5x+3)^2=2^2*(5x+3)^2$

In general, $\displaystyle (ab)^n=a^n * b^n$

So here, it's equal to $\displaystyle (2*(5x+3))^2$

6. Originally Posted by Moo
No, it's $\displaystyle 4*(5x+3)^2=2^2*(5x+3)^2$

In general, $\displaystyle (ab)^n=a^n * b^n$

So here, it's equal to $\displaystyle (2*(5x+3))^2$

Ahaa! You're a smart one! Thank you very much!

7. Try this:

$\displaystyle 4(5x+3)^2 = 9$

1. Divide both sides by 4

$\displaystyle (5x+3)^2 = \frac{9}{4}$

2. Take square root of both sides

$\displaystyle 5x + 3 = \pm\frac{3}{2}$

3. Solve

$\displaystyle 5x + 3 = \frac{3}{2}$ and $\displaystyle 5x + 3 = \frac{-3}{2}$

$\displaystyle x = \frac{-3}{10}$ and $\displaystyle x = \frac{-9}{10}$

8. Originally Posted by masters
Try this:

$\displaystyle 4(5x+3)^2 = 9$

1. Divide both sides by 4

$\displaystyle (5x+3)^2 = \frac{9}{4}$

2. Take square root of both sides

$\displaystyle 5x + 3 = \pm\frac{3}{2}$

3. Solve

$\displaystyle 5x + 3 = \frac{3}{2} and 5x + 3 = \frac{-3}{2}$

$\displaystyle x = \frac{-3}{10} and x = \frac{-9}{10}$

That was one other good way to solve it, maybe a bit simpler... Wonder wich I should show my teacher. Well i could just show them both

Thank you very much for the help!

9. Go for it. Be a rebel.

10. Originally Posted by liljeros
That was one other good way to solve it, maybe a bit simpler... Wonder wich I should show my teacher. Well i could just show them both

Thank you very much for the help!
It depends on what you're studying now
Or make a survey in your class to see which one is the most original one