How do you solve #7+sqrt(8-x)=12# algebraically and graphically?

Question

2004 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-03T13:39:16

The solution is #x=-17#

The equation is

#7+sqrt(8-x)=12#

#=>#, #sqrt(8-x)=12-7=5#

Squaring both sides

#(sqrt(8-x))^2=5^2#

#=>#, #8-x=25#

#=>#, ##x=8-25=-17

Graphically, plot the curve

#sqrt(8-x)+7-12=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)#

Answer 2 · 2018-07-03T13:50:46

I tried this:

1) Algebraically:
rearrange:

#sqrt(8-x)=12-7#

#sqrt(8-x)=5#

square both:

#(sqrt(8-x))^2=5^2#

#8-x=25#

#x=-25+8=-17#

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