A number added to itself six additional times is four less than the number squared.

Question

A.
2003-05-03-00-00_files/i0220000.jpg
B.
2003-05-03-00-00_files/i0220001.jpg
C.
2003-05-03-00-00_files/i0220002.jpg
D.
2003-05-03-00-00_files/i0220003.jpg

Answer 1

I'm pretty sure that a number added to itself 6 additional times would be x + 6x = 7x

 

We have:  7x=x²-4

Rearranging to form ax²+bx+c=0  gives us  x²-7x-4=0

 

With the quadratic x=(-b±√(b²-4ac))/2a  we have a=1, b=-7, c=-4

 

x= (7±√((-7)²-4(1)(-4))/2(1) = (7±√(49+16))/2 = (7±√65)/2

 

x = (7+√65)/2  and x = (7 - √65)/2 

Answer 2

Hi Lakeisha,

I'm not sure what to do with the files you uploaded, since they're just text to us and not links. BUT I can help you solve the problem.

What you need to do is set up an equation that says the number (n) times 6 (which is the same as adding it to itself 6 times but way easier to type) is equal to n²-4: 6n=n²-4

Now, put everything on one side of the equals sign: n²-6n-4=0

Use the quadratic formula to solve for n.