SOLUTION: place the following equations in the standard form of a quadratic equation (ax^2+bx+c=0) a. 4/x^2+3/x-10=0 b. 6+4/x-1=5/x-3 c. 3.5x+4.6/x=1.2 please help

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: place the following equations in the standard form of a quadratic equation (ax^2+bx+c=0) a. 4/x^2+3/x-10=0 b. 6+4/x-1=5/x-3 c. 3.5x+4.6/x=1.2 please help      Log On


   

Question 6686: place the following equations in the standard form of a quadratic equation (ax^2+bx+c=0)
a. 4/x^2+3/x-10=0
b. 6+4/x-1=5/x-3
c. 3.5x+4.6/x=1.2

please help
Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
I'll solve this one and you try other ones (same idea)
a. 4/x^2+3/x-10=0
use y instead of 1/x.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case 4y%5E2%2B3y%2B-10+=+0) has the following solutons:

y%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=%283%29%5E2-4%2A4%2A-10=169.

Discriminant d=169 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+169+%29%29%2F2%5Ca.

y%5B1%5D+=+%28-%283%29%2Bsqrt%28+169+%29%29%2F2%5C4+=+1.25
y%5B2%5D+=+%28-%283%29-sqrt%28+169+%29%29%2F2%5C4+=+-2

Quadratic expression 4y%5E2%2B3y%2B-10 can be factored:
4y%5E2%2B3y%2B-10+=+4%28y-1.25%29%2A%28y--2%29
Again, the answer is: 1.25, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B3%2Ax%2B-10+%29


so, my solver says y=5/4, -2. That means that x=1/y = 4/5, -1/2.