SOLUTION: Solve equations for x (x-3)^2=10 (x+2)^2 = 16

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Question 103616: Solve equations for x

(x-3)^2=10
(x+2)^2 = 16
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
1.
%28x-3%29%5E2=10
x%5E2+-6x+%2B9+=+10
x%5E2+-6x+%2B9+-+10+=+0
x%5E2+-+6x+-1+=+0

Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve x%5E2-6%2Ax-1=0 ( notice a=1, b=-6, and c=-1)





x+=+%28--6+%2B-+sqrt%28+%28-6%29%5E2-4%2A1%2A-1+%29%29%2F%282%2A1%29 Plug in a=1, b=-6, and c=-1




x+=+%286+%2B-+sqrt%28+%28-6%29%5E2-4%2A1%2A-1+%29%29%2F%282%2A1%29 Negate -6 to get 6




x+=+%286+%2B-+sqrt%28+36-4%2A1%2A-1+%29%29%2F%282%2A1%29 Square -6 to get 36 (note: remember when you square -6, you must square the negative as well. This is because %28-6%29%5E2=-6%2A-6=36.)




x+=+%286+%2B-+sqrt%28+36%2B4+%29%29%2F%282%2A1%29 Multiply -4%2A-1%2A1 to get 4




x+=+%286+%2B-+sqrt%28+40+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%286+%2B-+2%2Asqrt%2810%29%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%286+%2B-+2%2Asqrt%2810%29%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


x+=+%286+%2B+2%2Asqrt%2810%29%29%2F2 or x+=+%286+-+2%2Asqrt%2810%29%29%2F2



Now break up the fraction



x=%2B6%2F2%2B2%2Asqrt%2810%29%2F2 or x=%2B6%2F2-2%2Asqrt%2810%29%2F2



Simplify



x=3%2Bsqrt%2810%29 or x=3-sqrt%2810%29



So the solutions are:

x=3%2Bsqrt%2810%29 or x=3-sqrt%2810%29




2.
%28x%2B2%29%5E2+=+16
x%5E2+%2B+4x+%2B+4+=+16
x%5E2+%2B+4x+%2B+4+-+16+=+0+
x%5E2+%2B+4x+-12+=+0
Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve x%5E2%2B4%2Ax-12=0 ( notice a=1, b=4, and c=-12)





x+=+%28-4+%2B-+sqrt%28+%284%29%5E2-4%2A1%2A-12+%29%29%2F%282%2A1%29 Plug in a=1, b=4, and c=-12




x+=+%28-4+%2B-+sqrt%28+16-4%2A1%2A-12+%29%29%2F%282%2A1%29 Square 4 to get 16




x+=+%28-4+%2B-+sqrt%28+16%2B48+%29%29%2F%282%2A1%29 Multiply -4%2A-12%2A1 to get 48




x+=+%28-4+%2B-+sqrt%28+64+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-4+%2B-+8%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-4+%2B-+8%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


x+=+%28-4+%2B+8%29%2F2 or x+=+%28-4+-+8%29%2F2


Lets look at the first part:


x=%28-4+%2B+8%29%2F2


x=4%2F2 Add the terms in the numerator

x=2 Divide


So one answer is

x=2




Now lets look at the second part:


x=%28-4+-+8%29%2F2


x=-12%2F2 Subtract the terms in the numerator

x=-6 Divide


So another answer is

x=-6


So our solutions are:

x=2 or x=-6