(x-3)^2+(x+2)^2=17
How do I solve for x?
expand the two binomials, combine like terms, then put it in the form of a quadratic...you can then use the quadratic equation, or perhaps factor to find x.
here is what i got:
(x-3)(x-3)+(x+2)(x=2=17
2x^2-2x=22
x=11
Does that seem right?
(x-3)^2+(x+2)^2=17
x^2 -6 x + 9 + x^2 + 4 x + 4 -17 = 0
2 x^2 -2 x -4 = 0
(2x-4)(x+1) = 0
x = 4/2 = 2
or
x = -1
thank you very much for all your help, you really help by showing how to solve for x rather than just tell me what to do. I have a better understanding.
To solve the equation (x-3)^2 + (x+2)^2 = 17, you can follow these steps:
Step 1: Expand the expressions
(x-3)^2 = (x-3)(x-3) = x^2 - 6x + 9
(x+2)^2 = (x+2)(x+2) = x^2 + 4x + 4
Step 2: Substitute the expanded expressions into the original equation
(x^2 - 6x + 9) + (x^2 + 4x + 4) = 17
Step 3: Simplify the equation
Combine like terms to get:
2x^2 - 2x + 13 = 17
Step 4: Rearrange the equation to form a quadratic equation
Subtract 17 from both sides of the equation:
2x^2 - 2x - 4 = 0
Step 5: Divide the equation by 2 to simplify the coefficients:
x^2 - x - 2 = 0
Step 6: Factor the quadratic equation (if possible) or use the quadratic formula to solve for x. In this case, the equation can be factored as:
(x - 2)(x + 1) = 0
Step 7: Set each factor equal to zero and solve for x:
x - 2 = 0 or x + 1 = 0
Solving for x in each equation gives:
x = 2 or x = -1
Therefore, the solutions to the equation (x-3)^2 + (x+2)^2 = 17 are x = 2 and x = -1.