SOLUTION: 8. Solve and Check the Following Equations: A. 2^(x^2+x) = 64

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Question 113823: 8. Solve and Check the Following Equations:
A. 2^(x^2+x) = 64
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
2%5E%28x%5E2%2Bx%29=64 Start with the given equation


2%5E%28x%5E2%2Bx%29=2%5E6 Rewrite 64 as 2%5E6


Since the bases are equal, the exponents are equal. So x%5E2%2Bx=6

x%5E2%2Bx=6 Now let's solve for x


x%5E2%2Bx-6=0 Subtract x from both sides



%28x%2B3%29%28x-2%29=0 Factor the left side (note: if you need help with factoring, check out this
solver)



Now set each factor equal to zero:

x%2B3=0 or x-2=0

x=-3 or x=2 Now solve for x in each case


So our solutions are x=-3 or x=2


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Check:
Let's check the solution x=-3

2%5E%28x%5E2%2Bx%29=64 Start with the given equation


2%5E%28%28-3%29%5E2%2B%28-3%29%29=64 Plug in x=-3


2%5E%289%2B%28-3%29%29=64 Square -3 to get 9


2%5E6=64 Add

64=64 Raise 2 to the sixth power to get 64. Since both sides of the equation are equal, this solution is verified.

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Let's check the solution x=2

2%5E%28x%5E2%2Bx%29=64 Start with the given equation


2%5E%28%282%29%5E2%2B%282%29%29=64 Plug in x=2


2%5E%289%2B%282%29%29=64 Square 2 to get 4


2%5E6=64 Add

64=64 Raise 2 to the sixth power to get 64. Since both sides of the equation are equal, this solution is verified.


So this verifies our solutions x=-3 and x=2