SOLUTION: Solve by completeing square x^2=4-7x 3x^2-x-x-2=0 If the square of 2 less than an interger is 16, find the number(s)?

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Question 103617: Solve by completeing square
x^2=4-7x
3x^2-x-x-2=0
If the square of 2 less than an interger is 16, find the number(s)?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

1. x%5E2=4-7x => x%5E2+%2B+7x+-4+=+0

Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert 1x%5E2%2B7x%2B-4=0 to standard form by dividing both sides by 1:
We have: 1x%5E2%2B7x%2B-4=0. What we want to do now is to change this equation to a complete square %28x%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28x%2Bsomenumber%29%5E2: %28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1x%5E2%2Bhighlight_red%28+7%29+%2A+x%2B-4=0 that goes in front of x is 7, we know that 7=2*somenumber, or somenumber+=+7%2F2. So, we know that our equation can be rewritten as %28x%2B7%2F2%29%5E2+%2B+othernumber, and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that %28x%2B7%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B7x%2Bhighlight_green%28+-4+%29=0.


The highlighted red part must be equal to -4 (highlighted green part).

7%5E2%2F4+%2B+othernumber+=+-4, or othernumber+=+-4-7%5E2%2F4+=+-16.25.
So, the equation converts to %28x%2B7%2F2%29%5E2+%2B+-16.25+=+0, or %28x%2B7%2F2%29%5E2+=+16.25.

Our equation converted to a square %28x%2B7%2F2%29%5E2, equated to a number (16.25).

Since the right part 16.25 is greater than zero, there are two solutions:

system%28+%28x%2B7%2F2%29+=+%2Bsqrt%28+16.25+%29%2C+%28x%2B7%2F2%29+=+-sqrt%28+16.25+%29+%29
, or

system%28+%28x%2B7%2F2%29+=+4.03112887414927%2C+%28x%2B7%2F2%29+=+-4.03112887414927+%29
system%28+x%2B7%2F2+=+4.03112887414927%2C+x%2B7%2F2+=+-4.03112887414927+%29
system%28+x+=+4.03112887414927-7%2F2%2C+x+=+-4.03112887414927-7%2F2+%29

system%28+x+=+0.531128874149275%2C+x+=+-7.53112887414927+%29
Answer: x=0.531128874149275, -7.53112887414927.



2. 3x%5E2-x-x-2=0 => 3x%5E2+-2x+-+2+=+0

Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert 3x%5E2%2B-2x%2B-2=0 to standard form by dividing both sides by 3:
We have: 1x%5E2%2B-0.666666666666667x%2B-0.666666666666667=0. What we want to do now is to change this equation to a complete square %28x%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28x%2Bsomenumber%29%5E2: %28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1x%5E2%2Bhighlight_red%28+-0.666666666666667%29+%2A+x%2B-0.666666666666667=0 that goes in front of x is -0.666666666666667, we know that -0.666666666666667=2*somenumber, or somenumber+=+-0.666666666666667%2F2. So, we know that our equation can be rewritten as %28x%2B-0.666666666666667%2F2%29%5E2+%2B+othernumber, and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that %28x%2B-0.666666666666667%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B-0.666666666666667x%2Bhighlight_green%28+-0.666666666666667+%29=0.


The highlighted red part must be equal to -0.666666666666667 (highlighted green part).

-0.666666666666667%5E2%2F4+%2B+othernumber+=+-0.666666666666667, or othernumber+=+-0.666666666666667--0.666666666666667%5E2%2F4+=+-0.777777777777778.
So, the equation converts to %28x%2B-0.666666666666667%2F2%29%5E2+%2B+-0.777777777777778+=+0, or %28x%2B-0.666666666666667%2F2%29%5E2+=+0.777777777777778.

Our equation converted to a square %28x%2B-0.666666666666667%2F2%29%5E2, equated to a number (0.777777777777778).

Since the right part 0.777777777777778 is greater than zero, there are two solutions:


, or





system%28+x+=+1.21525043702153%2C+x+=+-0.548583770354863+%29
Answer: x=1.21525043702153, -0.548583770354863.