SOLUTION: please help me with this problem? directions: for each quadratic function, find the x-intercepts, if they exist. Problem: f(x)= -2x^2-6x+5

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: please help me with this problem? directions: for each quadratic function, find the x-intercepts, if they exist. Problem: f(x)= -2x^2-6x+5      Log On


   

Question 223983: please help me with this problem?
directions: for each quadratic function, find the x-intercepts, if they exist.
Problem:
f(x)= -2x^2-6x+5
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=+-2x%5E2-6x%2B5

Step 1. To find the x-intercepts,set the equation equal to zero (that is f(x)=y=0}}}

-2x%5E2-6x%2B5=0


Step 2. To solve, you can use the quadratic formula given as

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

where a=-2, b=-6 and c=5.

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -2x%5E2%2B-6x%2B5+=+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=%28-6%29%5E2-4%2A-2%2A5=76.

Discriminant d=76 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+76+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+76+%29%29%2F2%5C-2+=+-3.67944947177034
x%5B2%5D+=+%28-%28-6%29-sqrt%28+76+%29%29%2F2%5C-2+=+0.679449471770337

Quadratic expression -2x%5E2%2B-6x%2B5 can be factored:
-2x%5E2%2B-6x%2B5+=+-2%28x--3.67944947177034%29%2A%28x-0.679449471770337%29
Again, the answer is: -3.67944947177034, 0.679449471770337. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-2%2Ax%5E2%2B-6%2Ax%2B5+%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