Question

How do you solve `4sqrtx=sqrt( 9x+9)`?

Answer 1

`x = 9/7`

Explanation:

First, square both sides:
`(4sqrtx)^color(blue)2 = (sqrt(9x+9))^color(blue)2`

`16x = 9x+9`

Subtract `color(blue)9x` from both sides of the equation:
`16x quadcolor(blue)(-quad9x) = 9x + 9 quadcolor(blue)(-quad9x)`

`7x = 9`

Finally, divide both sides by `7`:
`(7x)/color(blue)7 = 9/color(blue)7`

So the answer is:
`x = 9/7`

However, we should check our work by plugging our solution back into the original equation:
`4sqrt(9/7) = sqrt(9(9/7) + 9)`

`4*3/sqrt7 = sqrt(81/7 + 63/7)`

`12/sqrt7 = sqrt(144/7)`

`12/sqrt7 = 12/sqrt17`

Both sides of the equation equal, meaning that `x = 9/7` is true.

Hope this helps!

Answer 2

`x = 9/7`.

Explanation:

First of All,

`color(white)(xxx)4sqrt(x) = sqrt(9x + 9)`

`rArr 16x = 9x + 9` [Square both sides]

`rArr 16x - 9x = 9` [Transpose `9x` to L.H.S.]

`rArr 7x = 9`

`rArr x = 9/7`.

So, `x = 9/7`.

But if we substitute `x = 9/7` in the equation, we get,

L.H.S. = `4sqrt(9/7) = sqrt(16 * 9/7) = sqrt(144/7)`

R.H.S = `sqrt(9*9/7 + 9) = sqrt(81/7 + 9) = sqrt((81 + 63)/7) = sqrt(144/7)`

So, L.H.S. = R.H.S.

So `x = 9/7` is indeed a solution for this equation.

Hope this helps.