gaby131o

2022-09-20

Consider the following function

$$r(x)=-(x+5{)}^{2}+2$$

$$r(x)=-(x+5{)}^{2}+2$$

Cremolinoer

Beginner2022-09-21Added 11 answers

Given $$r(x)=-(x+5{)}^{2}+2$$

so

$$0=-(x+5{)}^{2}+2\phantom{\rule{0ex}{0ex}}-(x+5{)}^{2}+2=0\phantom{\rule{0ex}{0ex}}-({x}^{2}+2x\times 5+{5}^{2})+2=0\phantom{\rule{0ex}{0ex}}-{x}^{2}-10x-25+2=0\phantom{\rule{0ex}{0ex}}-{x}^{2}-10-23=0\phantom{\rule{0ex}{0ex}}{x}^{2}+10x+23=0$$

Now we can solve this equation using quadratic formula

$$x=\frac{-10\pm \sqrt{{10}^{2}-4\times 23}}{2\times 1}\phantom{\rule{0ex}{0ex}}x=-10\pm \frac{\sqrt{100-92}}{2}\phantom{\rule{0ex}{0ex}}x=-10\pm \frac{\sqrt{8}}{2}\phantom{\rule{0ex}{0ex}}x=-10\pm \frac{\sqrt{4\times 2}}{2}\phantom{\rule{0ex}{0ex}}x=2(\frac{-5\pm \sqrt{2}}{2})\phantom{\rule{0ex}{0ex}}x=-5\pm \sqrt{2}$$

so

$$0=-(x+5{)}^{2}+2\phantom{\rule{0ex}{0ex}}-(x+5{)}^{2}+2=0\phantom{\rule{0ex}{0ex}}-({x}^{2}+2x\times 5+{5}^{2})+2=0\phantom{\rule{0ex}{0ex}}-{x}^{2}-10x-25+2=0\phantom{\rule{0ex}{0ex}}-{x}^{2}-10-23=0\phantom{\rule{0ex}{0ex}}{x}^{2}+10x+23=0$$

Now we can solve this equation using quadratic formula

$$x=\frac{-10\pm \sqrt{{10}^{2}-4\times 23}}{2\times 1}\phantom{\rule{0ex}{0ex}}x=-10\pm \frac{\sqrt{100-92}}{2}\phantom{\rule{0ex}{0ex}}x=-10\pm \frac{\sqrt{8}}{2}\phantom{\rule{0ex}{0ex}}x=-10\pm \frac{\sqrt{4\times 2}}{2}\phantom{\rule{0ex}{0ex}}x=2(\frac{-5\pm \sqrt{2}}{2})\phantom{\rule{0ex}{0ex}}x=-5\pm \sqrt{2}$$

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