SOLUTION: {{{sqrt(2x-1)-sqrt(x-5)=3}}} I need help showing the work for solving for X. I know that X=5, X=41. Thank you!

Algebra ->  Inequalities -> SOLUTION: {{{sqrt(2x-1)-sqrt(x-5)=3}}} I need help showing the work for solving for X. I know that X=5, X=41. Thank you!      Log On


   

Question 215271: sqrt%282x-1%29-sqrt%28x-5%29=3
I need help showing the work for solving for X. I know that X=5, X=41.
Thank you!
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282x-1%29-sqrt%28x-5%29=3


Step 1. Square both sides

sqrt%282x-1%29-sqrt%28x-5%29=3

%28sqrt%282x-1%29-sqrt%28x-5%29%29%5E2=3%5E2

2x-1+-+2%2Asqrt%282x-1%29%2Asqrt%28x-5%29%2Bx-5=9

3x-6-2%2Asqrt%28%282x-1%29%2A%28x-5%29%29=9

-2sqrt%28%282x-1%29%28x-5%29%29=9-3x%2B6=15-3x

2sqrt%282x%5E2-11x%2B5%29=3x-15

Step 2. Square both sides of equation

4%282x%5E2-11x%2B5%29=9x%5E2-90x%2B225

8x%5E2-44x%2B20=9x%5E2-90x%2B225

x%5E2-46x%2B205=0

Step 3. Use 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=-46 and c=205

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-46x%2B205+=+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=%28-46%29%5E2-4%2A1%2A205=1296.

Discriminant d=1296 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--46%2B-sqrt%28+1296+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-46%29%2Bsqrt%28+1296+%29%29%2F2%5C1+=+41
x%5B2%5D+=+%28-%28-46%29-sqrt%28+1296+%29%29%2F2%5C1+=+5

Quadratic expression 1x%5E2%2B-46x%2B205 can be factored:
1x%5E2%2B-46x%2B205+=+1%28x-41%29%2A%28x-5%29
Again, the answer is: 41, 5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-46%2Ax%2B205+%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