SOLUTION: Let f(x) = (x – 2)^2. Find x such that f(x) = 25 I do not understand the question :(

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Question 224605: Let f(x) = (x – 2)^2. Find x such that f(x) = 25
I do not understand the question :(
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Let f(x) = (x – 2)^2. Find x such that f(x) = 25

Step 1. So the equation becomes f%28x%29=25=%28x-2%29%5E2

Step 2. Multiply out the squared term: 25=x%5E2-4x%2B4

Step 3. Subtract 25 from both sides of the equation: 25-25=x%5E2-4x%2B4-25 or 0=x%5E2-4x-21

Step 4. Top solve, 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=1, b=-4, and c=-21

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-4x%2B-21+=+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=%28-4%29%5E2-4%2A1%2A-21=100.

Discriminant d=100 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+100+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+100+%29%29%2F2%5C1+=+7
x%5B2%5D+=+%28-%28-4%29-sqrt%28+100+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B-4x%2B-21 can be factored:
1x%5E2%2B-4x%2B-21+=+1%28x-7%29%2A%28x--3%29
Again, the answer is: 7, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B-21+%29



Step 5. Check solutions 7 and -3 with f%28x%29=25=%28x-2%29%5E2

f%287%29=25=%287-2%29%5E2=25 which is a true statement.

f%28-3%29=25=%28-3-2%29%5E2=25 which is another true statement.

Step 6. ANSWER: Solutions are x=7 and x=-3.

I hope the above steps and explanation 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