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Solve (x + 2)^2 + 7(x + 2) = 18. Identify any double roots. A. {–11, 0} B. {–2, –7} C. {–9, 2 (double root)} D. {–2, 9 (double root)}
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Do I simplify first or?...
Yeah simplify by finding that \[(x+2)^{2} = x ^{2}+4x+4\] Multiply out the 7(x+2) part to get 7x+14 add like terms so you get \[x ^{2}+4x+4+7x+4 = 18\] You get\[x ^{2}+11x+18=18\] Subtract 18 from both sides:\[x ^{2}+11x = 0\]Factor an x out you find you get \[x(x+11) = 0\] set both x = 0 and (x+11) = 0 to find x = 0 and x = -11 Answer is: A.
Thank you so much!!!
No problem.
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