SOLUTION: Solve. (x - 3)^2 + (x + 2)^2 = 17 I am not sure what to do after I distribute the square root through out the problem... This is what I have so far: (or is that wrong??) (x^2

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solve. (x - 3)^2 + (x + 2)^2 = 17 I am not sure what to do after I distribute the square root through out the problem... This is what I have so far: (or is that wrong??) (x^2       Log On


   



Question 391195: Solve.
(x - 3)^2 + (x + 2)^2 = 17
I am not sure what to do after I distribute the square root through out the problem... This is what I have so far: (or is that wrong??)
(x^2 + 9) + (x^2 + 4) = 17
x^2 + x^2 + 13 = 17

Found 2 solutions by haileytucki, richard1234:
Answer by haileytucki(390) About Me  (Show Source):
You can put this solution on YOUR website!
(x-3)^(2)+(x+2)^(2)=17
To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.
(x-3)^(2)+(x+2)^(2)-17=0
Squaring an expression is the same as multiplying the expression by itself 2 times.
(x-3)(x-3)+(x+2)^(2)-17=0
Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.
(x*x+x*-3-3*x-3*-3)+(x+2)^(2)-17=0
Simplify the FOIL expression by multiplying and combining all like terms.
(x^(2)-6x+9)+(x+2)^(2)-17=0
Squaring an expression is the same as multiplying the expression by itself 2 times.
(x^(2)-6x+9)+(x+2)(x+2)-17=0
Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.
(x^(2)-6x+9)+(x*x+x*2+2*x+2*2)-17=0
Simplify the FOIL expression by multiplying and combining all like terms.
(x^(2)-6x+9)+(x^(2)+4x+4)-17=0
Remove the parentheses that are not needed from the expression.
x^(2)-6x+9+x^(2)+4x+4-17=0
Since x^(2) and x^(2) are like terms, add x^(2) to x^(2) to get 2x^(2).
2x^(2)-6x+9+4x+4-17=0
Since -6x and 4x are like terms, subtract 4x from -6x to get -2x.
2x^(2)-2x+9+4-17=0
Add 4 to 9 to get 13.
2x^(2)-2x+13-17=0
Subtract 17 from 13 to get -4.
2x^(2)-2x-4=0
Factor out the GCF of 2 from each term in the polynomial.
2(x^(2))+2(-x)+2(-2)=0
Factor out the GCF of 2 from 2x^(2)-2x-4.
2(x^(2)-x-2)=0
In this problem 1*-2=-2 and 1-2=-1, so insert 1 as the right hand term of one factor and -2 as the right-hand term of the other factor.
2(x+1)(x-2)=0
Divide both sides of the equation by 2. Dividing 0 by any non-zero number is 0.
(x+1)(x-2)=0
Set each of the factors of the left-hand side of the equation equal to 0.
x+1=0_x-2=0
Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.
x=-1_x-2=0
Set each of the factors of the left-hand side of the equation equal to 0.
x=-1_x-2=0
Since -2 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 2 to both sides.
x=-1_x=2
The complete solution is the set of the individual solutions.
x=-1,2

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
You forgot the x terms in the expansion. It should look like:

%28x%5E2+-+6x+%2B+9%29+%2B+%28x%5E2+%2B+4x+%2B+4%29+=+17

After that, you can collect like terms and solve via the quadratic formula.