SOLUTION: 25y^2+20y+4=0

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Question 225834: 25y^2+20y+4=0
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
25y%5E2%2B20y%2B4=0

Step 1. Use quadratic formula given below to solve problem.

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a=25, b=20 and c=4.

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 25x%5E2%2B20x%2B4+=+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=%2820%29%5E2-4%2A25%2A4=0.

Discriminant d=0 is zero! That means that there is only one solution: x+=+%28-%2820%29%29%2F2%5C25.
Expression can be factored: 25x%5E2%2B20x%2B4+=+25%28x--0.4%29%2A%28x--0.4%29

Again, the answer is: -0.4, -0.4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+25%2Ax%5E2%2B20%2Ax%2B4+%29



Step 2. ANSWER: Solutions are -0.4 and -0.4

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J
http://www.FreedomUniversity.TV