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mat 117 dq2 week 7

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The rst thing we have to do is to put the loose number on the other side of the equaon. In our case
this is already done:
x
2
+10x=2
Next we must make operaons in order that the coe!cient of x
2
is 1 (this is also already done in our
case).
Now we look at the coe!cient of x. We divide this coe!cient by 2 (10/2=5) and the result we square
(5
2
=25). Now this last value will be added on both sides of the equaon. So:
x
2
+10x=2
|+25
x
2
+10x +25=2+25
x
2
+10x +25=27
Now this process creates on one side a perfect square:
x
2
+10x +25=27
(x+5)
2
=27
Now we can square-root both sides of the equaons and eventually solve the equaon:
(x +5)
2
=
27
x+5=
±3
3
|-5
x= -5
±3
3
And this is all.

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The first thing we have to do is to put the loose number on the other side of the equation. In our case this is already done: Next we must make operations in order that the coefficient of x2 is 1 (this is also already done in our case). Now we look at the coefficient of x. We divide this coefficie ...
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