I need help with this problem not understanding it. Give exact and approximate solutions to three decimal places. x^2+6x+9= 81 what are the exact solutions? x=___ There are no approximate solutions since the solutions are integers.
x^2 + 6 x - 72 = 0
(x-6)(x+12) = 0
x = 6 or x = -12
To find the exact solutions to the equation x^2 + 6x + 9 = 81, we can start by rearranging the equation to create a quadratic equation:
x^2 + 6x + 9 - 81 = 0.
Combine like terms:
x^2 + 6x - 72 = 0.
Now that we have a quadratic equation in the form ax^2 + bx + c = 0, we can solve for x by factoring, completing the square, or using the quadratic formula.
In this case, let's solve by factoring. We are looking for two numbers that multiply to -72 and add up to 6. The numbers that satisfy these conditions are -6 and 12.
Therefore, the factored form of the equation becomes:
(x - 6)(x + 12) = 0.
Now we can set each factor equal to zero and solve for x:
x - 6 = 0 or x + 12 = 0.
Solving each equation gives us:
x = 6 or x = -12.
So the exact solutions to the equation x^2 + 6x + 9 = 81 are x = 6 and x = -12.