SOLUTION: solve the given equation bu completing the suqare and applying the square root property y^2-8y=-7 y=? or y=?

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Question 212975: solve the given equation bu completing the suqare and applying the square root property
y^2-8y=-7
y=? or y=?
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the given equation by completing the square and applying the square root property

y%5E2-8y=-7

y=? or y=?

Step 1. Add 16 to both sides of the equation to get a perfect square on the left. Why 16? I divided -8 by 2 which gives -4 and then squared -4 to get 16.

y%5E2-8y%2B4%5E2=-7%2B4%5E2

%28y-4%29%5E2+=+9

Step 2. Taking the square root on both sides

sqrt%28%28y-4%29%5E2%29+=+sqrt%289%29

y-4=+3 and y-4=+-3

Solving yields

y-4%2B4=3%2B4

y=7 ANSWER

Step 3. The other solution is found as follows:

y-4=+-3

y-4%2B4+=+-3%2B4

y=1 ANSWER

Step 4. ANSWER IS x=7 and x=1

Step 5. As a check, the graph of the quadratic equation is shown below and note at x=7 and x=1 intersect the axis when y=0. So answer is correct

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-8x%2B7+=+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=%28-8%29%5E2-4%2A1%2A7=36.

Discriminant d=36 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--8%2B-sqrt%28+36+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-8%29%2Bsqrt%28+36+%29%29%2F2%5C1+=+7
x%5B2%5D+=+%28-%28-8%29-sqrt%28+36+%29%29%2F2%5C1+=+1

Quadratic expression 1x%5E2%2B-8x%2B7 can be factored:
1x%5E2%2B-8x%2B7+=+1%28x-7%29%2A%28x-1%29
Again, the answer is: 7, 1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-8%2Ax%2B7+%29



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.

Good luck in your studies!

Respectfully,
Dr J