Question

How do you solve `\sqrt { 5x + 95} = 3x + 7`?

Answer 1

Restrict the domain to `{x in RR|3x+7>=0}`
Square both sides and solve the resulting quadratic.
Check your work.

Explanation:

Given: `sqrt(5x + 95) = 3x + 7`

Restrict the domain so that the argument of the radical is greater than or equal to 0:

`5x+95 >=0`

`5x >=-95`

`x >=-19`

Another restriction would be that the result of the square root be greater than or equal to 0:

`3x+7>=0`

`x >= -7/3`

`sqrt(5x + 95) = 3x + 7; x>=-7/3`

Square both sides of the equation:

`5x+95= 9x^2 + 42x + 49; x >=-7/3`

Subtract `5x + 95` from both sides:

`9x^2 + 37x -46; x >=-7/3`

It is easy to see that 1 is a root, therefore, `(x - 1)` is a factor; this makes the other factor `(9x+46)`. The latter root violates the restriction:

`color(blue)(x = 1)`

Check

`sqrt(5(1) + 95) = 3(1) + 7`

`sqrt(100) = 10`

`10 = 10`

For those who would question the validity of discarding the second root, here is a graph of the curves `y = sqrt(5x+95)` and `y = 3x+7`

graph{(sqrt(5x+95)-y)(3x+7-y)=0 [-10, 10, -5, 15]}

Please observe that they intersect only once at the point `(1,10)`