If #x^2+16x+64=0#, what is the value of #x# ?

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Answer 1 · 2018-04-22T23:50:22

#x = -8#

Assuming you're trying to find #x#, you can start by factorizing the equation;

#x^2 + 16x + 64 = 0#

#(x+8)(x+8) = 0#

From the null factor law, either one of the products would have to equal #0#. In this case, it is #x+8#.

#x+8=0#

To solve for #x#, move other terms to leave #x# on its own, which makes

#x = -8#