Question

How do you solve `\sqrt{x}=x-6`?

Answer 1

`x = 9`

Explanation:

`sqrt(x) = x- 6`
Square the equation:
`x = (x-6)^2`
Apply the expansion of `(a- b)^2 = a^2 -2ab + b^2`
`implies x = x^2 - 12x + 36`
`implies 0 = x^2 - 13x + 36`
Factorize the quadratic.
`implies x^2 - 9x -4x + 36 = 0`
`implies x(x-9)-4(x-9) = 0`
`implies (x-4)(x-9) = 0`
`implies x = 4 or x = 9`

Note that substituting 4 in the equation returns 2 = -2, which is obviously wrong. So we neglect x = 4 in the set of solutions. Take care to verify your answers after solving(don't make my mistake!)

Answer 2

`x = 9`

Explanation:

`sqrtx = x - 6`

First, square both sides:
`sqrtx^color(red)(2) = (x-6)^color(red)2`

Simplify:
`x = x^2 - 12x + 36`

Move everything to one side of the equation:
`0 = x^2 - 13x + 36`

Now we need to factor.
Our equation is standard form, or `ax^2 + bx + c`.

The factored form is `(x-m)(x-n)`, where `m` and `n` are integers.

We have two rules to find `m` and `n`:

• `m` and `n` have to multiply up to `a * c`, or `36`
• `m` and `n` have to add up to `b`, or `-13`

Those two numbers are `-4` and `-9`. So we put them into our factored form:

`0 = (x-4)(x-9)`

Therefore,

`x - 4 = 0` and `x - 9 = 0`

`x = 4` `quadquadquad` and `quadquadquad` #``x = 9#

`--------------------`

However, we still need to check our answers by substituting them back into the original equation, since we have a square root in our original equation.

Let's first check if `x = 4` is really a solution:
`sqrt4 = 4 - 6`

`2 = -2`

This is not true! That means that `x !=4` (`4` is not a solution)

Now let's check `x = 9`:
`sqrt9 = 9 - 6`

`3 = 3`

This is true! That means that `x = 9` (`9` is really a solution)

So the final answer is `x = 9`.

Hope this helps!

Answer 3

`x=9` is the only real solution to this equation.

Explanation:

First, square both sides of this equation.

`x=x^2-12x+36`

Now put in standard form.

`x^2-13x+36=0`

Factor.

`(x-4)(x-9)=0`

`x=9` is a solution to this equation. `x=4` is not a solution to the original equation. However it is a solution to

`x=x^2-12x+36`

When we squared both sides to at the beginning, we enabled an extraneous solution since `(-sqrtx)^2=(sqrtx)^2=x`. Thus we enabled `-sqrtx` as a valid left-hand side of the equation when the original problem did not. Note that `-sqrtx=x-6` when `x=4`, but this is not what the problem is asking.