SOLUTION: I have to solve using the quadric formula and if there are no real roots say so...Can you please help me? Thanks...:) Page588 Number 14 3 = 7x - 4x^2

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I have to solve using the quadric formula and if there are no real roots say so...Can you please help me? Thanks...:) Page588 Number 14 3 = 7x - 4x^2      Log On


   

Question 146571This question is from textbook Beginning Algebra
: I have to solve using the quadric formula and if there are no real roots say so...Can you please help me? Thanks...:)
Page588
Number 14
3 = 7x - 4x^2 This question is from textbook Beginning Algebra

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
3 = 7x - 4x^2
Put into quadratic form (not quadric).
4x%5E2+-+7x+%2B+3+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 4x%5E2%2B-7x%2B3+=+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=%28-7%29%5E2-4%2A4%2A3=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--7%2B-sqrt%28+1+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-7%29%2Bsqrt%28+1+%29%29%2F2%5C4+=+1
x%5B2%5D+=+%28-%28-7%29-sqrt%28+1+%29%29%2F2%5C4+=+0.75

Quadratic expression 4x%5E2%2B-7x%2B3 can be factored:
4x%5E2%2B-7x%2B3+=+%28x-1%29%2A%28x-0.75%29
Again, the answer is: 1, 0.75. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B-7%2Ax%2B3+%29

The 2 roots (quadratics always have 2 roots) are real.