SOLUTION: Write a quadratic function having the given solutions: -5,3.

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Question 223995: Write a quadratic function having the given solutions: -5,3.
Found 2 solutions by solver91311, drj:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


If the zeros of the quadratic are -5 and 3, then so and so . That means the factors of the quadratic are and . Just multiply the two factors to get your quadratic.


John

Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Write a quadratic function having the given solutions: -5,3.

Step 1. With these solutions then x=-5 and x=3 or x+5=0 and x-3=0.

Step 2. Multiply x+5 and x-3 to get the quadratic equation using the FOIL method.

0=%28x%2B5%29%28x-3%29=x%5E2-3x%2B5x-15=x%5E2%2B2x-15

Step 3. ANSWER: With x=-5 and x=3, the corresponding quadratic equation is x%5E2%2B2x-15=0

We can check with the following:

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-15+=+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=%282%29%5E2-4%2A1%2A-15=64.

Discriminant d=64 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+64+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+64+%29%29%2F2%5C1+=+3
x%5B2%5D+=+%28-%282%29-sqrt%28+64+%29%29%2F2%5C1+=+-5

Quadratic expression 1x%5E2%2B2x%2B-15 can be factored:
1x%5E2%2B2x%2B-15+=+1%28x-3%29%2A%28x--5%29
Again, the answer is: 3, -5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-15+%29


I hope the above steps and explanation were helpful.

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

Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs.

Respectfully,
Dr J

http://www.FreedomUniversity.TV