SOLUTION: How does one solve this equation? (√10+(3√x)) = (√x) The radical over the 10 covers the whole equation on the left side of the equals sign.

Algebra ->  Radicals -> SOLUTION: How does one solve this equation? (√10+(3√x)) = (√x) The radical over the 10 covers the whole equation on the left side of the equals sign.       Log On


   

Question 842482: How does one solve this equation?
(√10+(3√x)) = (√x)
The radical over the 10 covers the whole equation on the left side of the equals sign.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
if i understand you correctly, your equation is:

sqrt(10 + 3 * sqrt(x)) = sqrt(x)

pushing this through the algebra.com equation generator make it look like this:

sqrt%2810+%2B+3+%2A+sqrt%28x%29%29+=+sqrt%28x%29

start with:
sqrt(10 + 3 * sqrt(x)) = sqrt(x)
square both sides of this equation to get:
10 + 3 * sqrt(x) = x
subtract 10 from both sides of the equation and subtract 3*sqrt(x) from both sides of the equation to get:
0 = x - 3*sqrt(x) - 10
commute this to get:
x - 3*sqrt(x) - 10 = 0
this is a quadratic equation in disguised form.
let y = sqrt(x) and this equation becomes:
y^2 - 3*y - 10 = 0
factor this equation to get:
(y-5) * (y+3) = 0
solve for y to get:
y = 5 or y = -3
replace y with sqrt(x) to get:
sqrt(x) = 5 or sqrt(x) = -3
solve for x to get:
x = 25 or x = 9
now you need to check to see if these solutions are good by replacing x in the original equation with these values of x to see if the equations hold true.

first we'll do 25.
the original equation is:
sqrt(10 + 3 * sqrt(x)) = sqrt(x)
replace x with 25 to get:
sqrt(10 + 3 * sqrt(25)) = sqrt(25)
simplify to get:
sqrt(10 + 3*5) = 5
simplify further to get:
sqrt(25) = 5
simplify further to get:
5 = 5
the equation is true so x = 25 is a good solution.

now we'll do 9.
the original equation is:
sqrt(10 + 3 * sqrt(x)) = sqrt(x)
replace x with 9 to get:
sqrt(10 + 3 * sqrt(9)) = sqrt(9)
simplify to get:
sqrt(10 + 3*3) = 3
simplify to get:
sqrt(19) = 3
this is not true, so x = -3 cannot be a good solution.

the problem is that the square root of (9) is not equal to -3.
it is equal to 3.
the square root of a number is always the positive root, unless you are solving for y^2.
example:
if your equation is y^2 = 9, then y = +/- sqrt(9) = +/- 3
sqrt(9) can be + 3 or it can be - 3 because you are solving for y^2.
if your equation is y = sqrt(9), however, then y = + 3 only.

the grpah of your 2 equations is shown below.
they are equal when the graph of the equations intersect.
the 2 equations to be graphed are:
y = sqrt(10 + 3*sqrt(x))
y = sqrt(x)

you can see that the graphs intersect at x = 25 and y = 5 which is the coordinate point of (25,5).