Solve the equation?

Question

3+square root x^2-8x=0

Answer 1

uhmmm do you mean to say 3+ sqrt (x^2-8x)=0
if so
sqrt(x^2 -8x) = -3
square both sides to get
x^2-8x = 9
then transpose back and FACTOR out
x^2-8x-9=0
(x-9)(x+1)=0
x is equal to 9 and -1

if you mean 3+ sqrt(x^2) - 8x=0
then sqrt of x^2 is equal to x and we have...
3+x -8x =0
3-7x=0
or x = 3/7

Answer 2

x^2 - 8x is inside the square root symbol?

If that's what you mean, then

3+sqrt (x^2-8x) = 0
sqrt (x^2-8x) = -3

From the above equation, it is safe to say that there are no solutions since the square root is defined to be the positive root.

Anyway, if you want to continue:
square both sides:
x^2 - 8x = 9
x^2 - 8x - 9 = 0
(x-9)(x+1) = 0
x-9 = 0 ||| x+1 =0
x = 9 ||| x = -1

Substitute x = 9 into the orig eqn:
3+sqrt (9^2-8(9)) = 0
3+sqrt (81-72) = 0
3+ sqrt(9) = 0
3 + 3 = 0 FALSE

Substitute x = -1 into the orig eqn:
3+sqrt ((-1)^2-8(-1)) = 0
3+sqrt (1+8) = 0
3+ sqrt(9) = 0
3 + 3 = 0 FALSE

The solution set is empty.

Answer 3

If it is 3+sqrt(x^2-8x)=0 No solution as both summands are positive
If it is
3+sqrt(x^2) -8x=0 then if x>=0
3+x-8x=0 7x=3 and x= 3/7
Ifx<0
3-x-8x=0 9x==3 x=1/3 Not a solution as x must be <0