Question

How do you solve `x= sqrt(6x) +7`?

Answer 1

`x=10+sqrt51`

Explanation:

`x=sqrt(6x)+7`
First let's define the domain :
`x>=0`
So:
`x-7=sqrt(6x)`
`(x-7)²=6x`
`(x²-14x+49)-6x=0`
`x²-20x+49=0`
`Δ=20²-4*49`
`Δ=204`
`x=(-b+-sqrtΔ)/2a`
`x_"1"=(20-sqrt204)/2`
`x_"2"=(20+sqrt204)/2`
`x_"1"=(20-2sqrt51)/2`
`x_"2"=(20+2sqrt51)/2`
`x_"1"=10-sqrt51<0=>` NOT a solution.
`x_"2"=10+sqrt51`
So : `x=10+sqrt51`
\0/ herex's our answer!

Answer 2

`x=10+sqrt51`

Bare with me please: this is a long solution process:

Explanation:

We can subtract `7` from both sides to arrive at:

`x-7=sqrt(6x)`

We can square both sides to get:

`color(blue)((x-7)^2)=6x`

`=color(blue)((x-7)(x-7))=6x`

What I have in blue, we can use the highly useful mnemonic FOIL to simplify this. We simply multiply the first terms, outside terms, inside and last terms. We get:

• First terms: `x*x=x^2`
• Outside terms: `x*-7=-7x`
• Inside terms: `-7*x=-7x`
• Last terms: `-7*-7=49`

Now, we have:

`color(blue)(x^2-7x-7x+49)=6x`

Which can be simplified to

`x^2-14x+49=6x`

We have a quadratic on the left, so we want to set it equal to zero to find its zeroes. We do this by subtracting `6x` from both sides to get:

`x^2-20x+49=0`

The only factors of `49` are `1, 7` and `49`, and neither of them, positive or negative, add up to `-20`. We can use the quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

where `a=1, b=-20 & c=49` (in the form `ax^2+bx+c=0`)

Plugging into the quadratic formula, we get:

`x=(20+-sqrt(400-4(1*49)))/(2*1)`

Which simplifies to

`x=(20+-color(blue)(sqrt(204)))/2`

Because `51*4=204`, we can rewrite `color(blue)(sqrt204)` as

`color(blue)(sqrt4*sqrt51)`

Which obviously simplifies to `color(blue)(2sqrt51)`

Now, we have

`x=(20+-2sqrt51)/2`

Every term is divisible by `2`, so we can factor a 2 out to get

`x=(color(red)(10)cancel(20)+-cancel2sqrt51)/(color(red)(1)cancel2)`

Which is equal to

`color(red)(x=10+-sqrt51)`

Since our answer had a square root in it, we know the domain has to be `x>=0`.

`10+sqrt51>0`, so this is a solution for `x` but

`10-sqrt51<0`, which is outside of the domain, so we get

`x=10+sqrt51`

as our final solution.

This was a long solution process, but all I did was:

• Get `x` on one side
• Square both sides to get rid of the square root
• Used FOIL to simplify the left side
• Used the Quadratic Formula
• Checked the domain

I really hope this helps!