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help me to get the solution
a.) (k+7)²=289
b.) (2s-1)²=225
c.) 3h²-147=0


Sagot :

[tex]a)\sqrt{ (k+7)^2}=\sqrt{289} \\\\ |k+7|=17 \\\\ 1)k+7=17 \\\\ k=17-7 \\\\ \boxed{k=10} \\\\\\ 2)k+7=-17 \\\\ k=-17-7 \\\\ \boxed{k=-24} \\\\\\ \boxed{\boxed{S=\{-24,10\}}}[/tex]

[tex]b)\sqrt{(2s-1)^2}=\sqrt{225} \\\\ |2s-1|=15 \\\\ 1)2s-1=15 \\\\ 2s=15+1 \\\\ 2s=16 \ \ \ |:2 \\\\ \boxed{s=8} \\\\\\ 2)2s-1=-15 \\\\ 2s=-15+1 \\\\ 2s=-14 \ \ \ |:2 \\\\ \boxed{s=-7} \\\\\\ \boxed{\boxed{S=\{-7;8\}}}[/tex]

[tex]c)3h^2-147=0 \ \ \ |:3 \\\\ h^2-49=0 \\\\ (h+7)(h-7)=0 \\\\ 1)h+7=0 \\\\ \boxed{h=-7} \\\\\\ 2)h-7=0 \\\\ \boxed{h=7} \\\\\\ \boxed{\boxed{S=\{\pm 7\}}}[/tex]