If x^2 = 16. what is the sum of all possible values of x?
Hello Guest!
\(x^2=16\) | take the root
\(x=\pm\sqrt{16}\) | calculate the root
\(x_1=4\\ x_2=-4\)
\(x_1+x_2=4+(-4)=0\)
\(The\ sum\ of\ all\ possible\ values\ of\ x\ is\ 0\ .\)
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If \(x^2 = 16\), then the possible values of x are \(\pm 4\), meaning 4 and -4.
\(4+-4=0\), so the sum of all possible values of x are 0.
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