SOLUTION: Hi, I've been trying for hours to figure out how to isolate y in the following equation. y(y-5)=546 I know y=26 but how do I prove/solve it? (I've come up with numerous vari

Algebra ->  Expressions -> SOLUTION: Hi, I've been trying for hours to figure out how to isolate y in the following equation. y(y-5)=546 I know y=26 but how do I prove/solve it? (I've come up with numerous vari      Log On


   

Question 9072: Hi, I've been trying for hours to figure out how to isolate y in the following equation.
y(y-5)=546
I know y=26 but how do I prove/solve it? (I've come up with numerous variations to the equation such as: 8y-40=4368/y, (8y-40)y=4368, y^2-5y=546, etc). Thanks in advance for your help.
Peter
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
y(y-5)=546
y^2 -5y -546 = 0
This gives us the roots 26 and -21.
The working is shown below.
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B-546+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-5%29%5E2-4%2A1%2A-546=2209.

Discriminant d=2209 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+2209+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+2209+%29%29%2F2%5C1+=+26
x%5B2%5D+=+%28-%28-5%29-sqrt%28+2209+%29%29%2F2%5C1+=+-21

Quadratic expression 1x%5E2%2B-5x%2B-546 can be factored:
1x%5E2%2B-5x%2B-546+=+%28x-26%29%2A%28x--21%29
Again, the answer is: 26, -21. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B-546+%29



Hope this helps,
Prabhat