The solution set is { }.
sq rt 2x+5=13
sq rt 2x=8
(sq rt 2x)^2=8 ^2
2x=64
x=32 2*32=64 sq rt 64=8 8+5=13
√2x + 5 = 13?
x=32
SqRt(2)x+5=13
sqrt(2)x=13-5
sqrt(2)x=8
x=8/sqrt(2)
x=4sqrt(2)=5.656
√2x + 5 = 13 ( subtract 5 from both sides)
√2x = 8 (square both sides)
2x = 64 (divide 2 from both sides)
x = 32
(This is if only the 2x is under the square root.)
√2x + 5 = 13 (square both sides)
2x+5 = 169 (subtract 5 from both sides)
2x = 164 (divide both sides by 2)
x = 82
(This is if both the 2x and the 5 in under the square root sign.)
root of 2x+5 = 13
then 2x+5 = 169
2x = 169 - 5 = 164
then x = 164/2 = 82
if (2x+5) is contained in the square root....
√2x + 5 = 13
square both sides
2x+5=169
2x=164
x=82
if..(2x) is the only term under the sqrt...
√2x = 13 - 5 =8
square both sides...
2x^1/2=8
x= ((13-5)^2)/2=32
x = 32, if the radical is over the x. x = 2^2.5 if not.
sqrt(2)*x + 5 = 13
sqrt(2)*x = 8
x = plusminus (8/sqrt(2))
So the solutions are either -5.657 or 5.657.
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Verified answer
sq rt 2x+5=13
sq rt 2x=8
(sq rt 2x)^2=8 ^2
2x=64
x=32 2*32=64 sq rt 64=8 8+5=13
√2x + 5 = 13?
x=32
SqRt(2)x+5=13
sqrt(2)x=13-5
sqrt(2)x=8
x=8/sqrt(2)
x=4sqrt(2)=5.656
√2x + 5 = 13 ( subtract 5 from both sides)
√2x = 8 (square both sides)
2x = 64 (divide 2 from both sides)
x = 32
(This is if only the 2x is under the square root.)
√2x + 5 = 13 (square both sides)
2x+5 = 169 (subtract 5 from both sides)
2x = 164 (divide both sides by 2)
x = 82
(This is if both the 2x and the 5 in under the square root sign.)
root of 2x+5 = 13
then 2x+5 = 169
2x = 169 - 5 = 164
then x = 164/2 = 82
if (2x+5) is contained in the square root....
√2x + 5 = 13
square both sides
2x+5=169
2x=164
x=82
if..(2x) is the only term under the sqrt...
√2x + 5 = 13
√2x = 13 - 5 =8
square both sides...
2x=64
x=32
2x^1/2=8
2x=64
x=32
x= ((13-5)^2)/2=32
x = 32, if the radical is over the x. x = 2^2.5 if not.
sqrt(2)*x + 5 = 13
sqrt(2)*x = 8
x = plusminus (8/sqrt(2))
So the solutions are either -5.657 or 5.657.