(x +3)^(2) + 10x \leqslant 21

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asked Sep 17, 2022 21.3k views

11 votes

$(x +3)^(2) + 10x \leqslant 21$

Mathematics
high-school

Nick Johnson asked

by Nick Johnson

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1 Answer

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Explanation:

${(x + 3)}^(2) + 10x\leqslant 10 \\ {x}^(2) + 6x + 9 + 10x \leqslant 10 \\ {x}^(2) + 16x + 9 \leqslant 10 \\ {x}^(2) + 6x + 9 \leqslant 0 \\ the \: value \: of \: x \: that \: makes \: it \: exist \\ x = - 3$

Jason Reeves answered Sep 21, 2022

by Jason Reeves

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