How do you simplify #sqrt( 8 x + 1) = 5#?

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Answer 1 · 2016-04-09T11:15:58

#x=3#

#color(blue)(sqrt( 8 x + 1) = 5#

Square both sides to get rid of the radical sign

#rarr(sqrt( 8 x + 1))^2 = 5^2#

#rarr8x+1=25#

#rarr8x=25-1#

#rarr8x=24#

#color(green)(rArrx=24/8=3#

Check

#color(brown)(sqrt( 8(3) + 1) = 5#

#color(brown)(sqrt(24+1)=5#

#color(brown)(sqrt25=5#

#color(green)(5=5#

Answer 2 · 2016-04-09T11:19:28

#x=3#

Square root is an 'action'. To 'get rid' of that action you reverse it!

The reverse of square root is to square it.

Square both sides giving:

#(sqrt(8x+1))^2=5^2#

#=>8x+1=25#

Subtract 1 from both sides (the revers of add is subtract)

#8x+1-1=25-1#

#8x=24#

Divide both sides by 8 (the reverse of multiply is divide).

#8/8xx x= 24/8#

But #8/8=1#

#x=3#