Question

How do you solve `\sqrt { x } - 4= \sqrt { 9x }`?

Answer 1

See a solution process below:

Explanation:

FIrst, add `color(red)(4)` and subtract `color(blue)(sqrt(9x))` from each side of the equation to isolate the radicals while keeping thee equation balanced:

`sqrt(x) - color(blue)(sqrt(9x)) - 4 + color(red)(4) = sqrt(9x) - color(blue)(sqrt(9x)) + color(red)(4)`

`sqrt(x) - color(blue)(sqrt(9x)) - 0 = 0 + color(red)(4)`

`sqrt(x) - color(blue)(sqrt(9x)) = color(red)(4)`

Next, square both sides of the equation to eliminate the radicals while keeping the equation balanced:

`(sqrt(x) - color(blue)(sqrt(9x)))^2 = color(red)(4)^2`

(Note: use this rule to multiply the left side of the equation:)
`(color(red)(x) - color(blue)(y))^2 = color(red)(x)^2 - 2color(red)(x)color(blue)(y) + color(blue)(y)^2`

`(sqrt(x))^2 - 2sqrt(x)color(blue)(sqrt(9x)) + (sqrt(9x))^2 = 16`

`x - 2sqrt(x * color(blue)(9x)) + 9x = 16`

`x - 2sqrt(color(blue)(9x^2)) + 9x = 16`

`x color(red)(+-) (2 * 3x) + 9x = 16`

`x color(red)(+-) 6x + 9x = 16`

Solution for `color(red)(-)` 6x:

Then, we can combine like terms:

`1x - 6x + 9x = 16`

`(1 - 6 + 9)x = 16`

`4x = 16`

Now, divide each side of the equation by `color(red)(4)` to solve for `x` while keeping the equation balanced:

`(4x)/color(red)(4) = 16/color(red)(4)`

`(color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4)) = 4`

`x = 4`

Solution for `color(red)(+)` 6x:

Then, we can combine like terms:

`1x + 6x + 9x = 16`

`(1 + 6 + 9)x = 16`

`16x = 16`

Now, divide each side of the equation by `color(red)(16)` to solve for `x` while keeping the equation balanced:

`(16x)/color(red)(16) = 16/color(red)(16)`

`(color(red)(cancel(color(black)(16)))x)/cancel(color(red)(16)) = 1`

`x = 1`

The Solutions Are: `x = 4` and `x = 1`

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(Note, if you substitute these into the original equation the are two things to notice:

1. The square root of a number produces both a positive and negative result.
2. When dealing with square roots there can be extraneous solutions.

Answer 2

No solution.

Explanation:

`sqrtx-4=sqrt(9x)`

Isolate the radicals by subtracting `sqrtx` from both sides.

`-4=sqrt(9x)-sqrtx`

Split `sqrt(9x)` into `sqrt9 * sqrtx` and simplify.

`-4=(sqrt9*sqrtx)-sqrtx`

`-4=3sqrtx-sqrtx`

Subtract the radicals.

`-4=2sqrtx`

Divide both sides by `2`.

`-2=sqrtx`

`sqrtx=-2`

Now, we are left with `sqrtx=-2`. Note that any value you think of for `x` will never yield a negative answer, because the square root of a number will never be negative. `x` cannot equal `4`, because `2 != -2`. `x` cannot equal `-4` because you cannot take `sqrt(-4)`. Thus, there is no solution.