Math algebra - Find the type & number of solutions for each equation

\({{x}^{2}}-6{x}=-5\)

Something like this?

As you can tell, it's a quadratic!

\({{x}^{2}}-6{x}+5=0\)

Now you solve for x:

\({{x}^{2}}-6{x}+5=0\)

Or, you can factor out an \({x}\):

\({x}({x}-6)=5
{x}=-5
{x}-6=-5
{x}=-5+6
{x}=1
\)

So our \({x}\) values are: -5 and 1

I will work on the rest for you.

I love this math module! Genius Idea

Woops I made a mistake with the factoring in the previous question

Yeah, I can see you're struggling, why don't you put it in vertex form and from there, solve for the x's?

Lets start with where you left off from, Star.

\(x2 - 6x = -5\)

x2 - 6x + 5 = 0

You need to find factors of \(5\) that add up to \(6\)... Let's try \(5 * 1 = 5\), but when we add them, we get \(+6\). So that won't work.

Since, it's not working, we have to convert it into vertex form by completing the square and then manually solving for \(x\).

\(x2 - 6x + 5 = 0\)

\(\frac{-6}{2}=(-3)^2=9

x2 - 6x + 9 - 9 + 5 = 0
(x-3)^2 - 4=0
\)

Now solve for \(x\):

\(\sqrt{4} = \sqrt{(x-3)^2}\)

+/-\(2 = x-3\)

if +2: 2 + 3 = \(x = 5\)
if -2: -2 + 3 = \(x = 1\)

Don't forget to write a "therefore" statement

Awesome, but we ended up getting the same answers , so I guess I was going it right afterall. Nini, are you ok with the other two or would u like me to answer them as well?

61. \(x^2-245x+144 = 0\)

\(x^2-245x +\)\(({\frac{245}{2})^{2}}\)\(-({\frac{245}{2})^{2}}\)\( +144 = 0\)

\((x - \frac{245}{2})^2 - 14862.25 = 0

+{sqrt{\14862.25}} + \frac{245}{2}= x = 244.41
-{sqrt{\14862.25}} + \frac{245}{2}= x = 0.58917
\)

SO FAR IM DOING GOOD WITH THE OTHERS I JUST NEED HELP WITH THE MORE ALGEBRA PART.   THANK YOU