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solve for k. (k+2)^2=3 express answer in simplest form
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\[This\ is\ of\ the\ form\ (a + b)^2\] \[k^2 + 4 + 4k = 3, k^2 + 4k = -1\] \[k^2 + 4k +1 = 0, k(k+4) +1 = 0\] \[(k+1)= 0 \ or\ (k+4) = 0, so \ k = -1 \ or\ k = -4\]
It is in its simplest form already. If you're trying to find the roots, here we go: \[k+2=\pm \sqrt{3}\] \[k=\pm \sqrt{3}-2\] There are two solutions.
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