Mathematics

1. (k+7)² =289
2. (2s-1)² =225
3. (x-4)² =169

please help me answer these :)


Answers

yassautumn

5 years ago Comment

[latex](k+7)^2 =289\\ |k+7|=17\\ k+7=17\vee k+7=-17\\ k=10 \vee k=-24[/latex]

[latex] (2s-1)^2 =225\\ |2s-1|=15\\ 2s-1=15\vee 2s-1=-15\\ 2s=16 \vee 2s=-14\\ s=8 \vee s=-7[/latex]

[latex](x-4)^2 =169\\ |x-4|=13\\ x-4=13 \vee x-4=-13\\ x=17 \vee x=-9[/latex]

Polieke

5 years ago Comment

[latex]1. \ \ \ (k+7)^2 =289\\\\\Leftrightarrow\ \ \ k+7= \sqrt{289} \ \ \ or\ \ \ \ k+7=- \sqrt{289}\\\\k=17-7=10\ \ \ \ \ \ \ or\ \ \ \ k=-17-7=-24 \\\\2.\ \ \ (2s-1)^2 =225\\\\\Leftrightarrow\ \ \ 2s-1= \sqrt{225} \ \ \ or\ \ \ \ 2s-1=- \sqrt{225}\\\\2s=15+1\ \ \ \ \ \ \ or\ \ \ \ 2s=-15+1\\2s=16\ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ 2s=-14\\s=8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ s=-7 \\\\[/latex]

[latex]3.\ \ \ (x-4)^2 =169\\\\\Leftrightarrow\ \ \ x-4= \sqrt{169} \ \ \ or\ \ \ \ x-4=- \sqrt{169}\\\\x=13+4=17\ \ \ \ \ \ \ or\ \ \ \ x=-13+4=-9[/latex]