How do you solve $7 + \sqrt{8 - x} = 12$ $7+sqrt(8-x)=12$ algebraically and graphically?

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How do you solve $7 + \sqrt{8 - x} = 12$ $7+sqrt(8-x)=12$ algebraically and graphically?

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Answer 1 · 2018-07-03T13:39:16

The solution is $x = - 17$ $x=-17$

Explanation:

The equation is

$7 + \sqrt{8 - x} = 12$ $7+sqrt(8-x)=12$

$\Rightarrow$ $=>$ , $\sqrt{8 - x} = 12 - 7 = 5$ $sqrt(8-x)=12-7=5$

Squaring both sides

${(\sqrt{8 - x})}^{2} = 5^{2}$ $(sqrt(8-x))^2=5^2$

$\Rightarrow$ $=>$ , $8 - x = 25$ $8-x=25$

$\Rightarrow$ $=>$ , x=8-25=-17

Graphically, plot the curve

$\sqrt{8 - x} + 7 - 12 = 0$ $sqrt(8-x)+7-12=0$

$\sqrt{8 - x} - 5 = 0$ $sqrt(8-x)-5=0$

graph{sqrt(8-x)-5 [-30.25, 10.3, -8.68, 11.58]}

And the solution to the equation is the point where the curve intercepts with the x-axis, that is $(- 17, 0)$ $(-17,0)$

Answer 2 · 2018-07-03T13:50:46

I tried this:

Explanation:

1) Algebraically:
rearrange:

$\sqrt{8 - x} = 12 - 7$ $sqrt(8-x)=12-7$

$\sqrt{8 - x} = 5$ $sqrt(8-x)=5$

square both:

${(\sqrt{8 - x})}^{2} = 5^{2}$ $(sqrt(8-x))^2=5^2$

$8 - x = 25$ $8-x=25$

$x = - 25 + 8 = - 17$ $x=-25+8=-17$

2) Graphically:
graph{(y-7-sqrt(8-x))(y-12)=0 [-50.02, 14.96, -3.34, 29.11]}

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How do you solve $7 + \sqrt{8 - x} = 12$ $7+sqrt(8-x)=12$ algebraically and graphically?

2 Answers

Explanation:

Explanation:

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