SOLUTION: How do I find the solution to this equation by completing the square: 2x^2-5x=7 Thanks for any help.

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Question 1047286: How do I find the solution to this equation by completing the square:
2x^2-5x=7
Thanks for any help.

Found 3 solutions by MathLover1, josgarithmetic, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

2x%5E2-5x=7
2x%5E2-5x-7=0 ....to complete square, write -5x as 2x-7x
2x%5E2%2B2x-7x-7=0....group
%282x%5E2%2B2x%29-%287x%2B7%29=0...factor out 2x and 7
2x%28x%2B1%29-7%28x%2B1%29=0
%282x-7%29%28x%2B1%29=0
so, the solution to this will be:
if %282x-7%29=0->2x=7->x=7%2F2->x=3.5
if %28x%2B1%29=0->x=-1

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+2x%5E2-5x-7%29+


Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Factorize the 2 on the left member.

2%28x%5E2-5x%2F2%29=7

The term to add to BOTH members to Complete The Square is %285%2F%282%2A2%29%29%5E2=%285%2F4%29%5E2.

CAREFUL! You do not simply add that to both members. WHY? Because you factored a 2 from the left member, but not the right member. You will not simply add (5/4)^2 to both sides. You add this term INSIDE the parentheses of the left member, .... and then you add 2%285%2F4%29%5E2 to the right member. THIS is the kind of thing you need to watch and understand to make progress.

That step is this:
2%28x%5E2-5x%2F2%2B%285%2F4%29%5E2%29=7%2B2%285%2F4%29%5E2
Be sure you understand.

You continue further steps toward solving x.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
How do I find the solution to this equation by completing the square:
2x^2-5x=7
Thanks for any help.
2x%5E2+-+5x+=+7
2x%5E2%2F2+-+%285%2F2%29x+=+%287%2F2%29 ------- DIVIDING each side by 2 to make the LEADING coefficient, 1
x%5E2+-+%285%2F2%29x+=+7%2F2
x%5E2+-+%285%2F2%29x+%2B+%28-+5%2F4%29%5E2+=+7%2F2+%2B+%28-+5%2F4%29%5E2 ------ Taking , SQUARING it, and ADDING this result to EACH SIDE
%28x+-+5%2F4%29%5E2+=+7%2F2+%2B+25%2F16
%28x+-+5%2F4%29%5E2+=+56%2F16+%2B+25%2F16 ------ Changing 7%2F2 to denominator, 16
%28x+-+5%2F4%29%5E2+=+81%2F16
sqrt%28%28x+-+5%2F4%29%29%5E2+=+%22+%22%2B-+sqrt%2881%2F16%29 ------- Taking square root of both sides
x+-+5%2F4+=+%22+%22%2B-+9%2F4
x+=+5%2F4+%2B-+9%2F4
OR