SOLUTION: Use the discriminat to determine how many real number solutions the quadratic equation -4j^2 + 3j-28=0 has

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Question 5545: Use the discriminat to determine how many real number solutions the quadratic equation -4j^2 + 3j-28=0 has
Answer by Abbey(339) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -4x%5E2%2B3x%2B28+=+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=%283%29%5E2-4%2A-4%2A28=457.

Discriminant d=457 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+457+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+457+%29%29%2F2%5C-4+=+-2.29719479080399
x%5B2%5D+=+%28-%283%29-sqrt%28+457+%29%29%2F2%5C-4+=+3.04719479080399

Quadratic expression -4x%5E2%2B3x%2B28 can be factored:
-4x%5E2%2B3x%2B28+=+-4%28x--2.29719479080399%29%2A%28x-3.04719479080399%29
Again, the answer is: -2.29719479080399, 3.04719479080399. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-4%2Ax%5E2%2B3%2Ax%2B28+%29