Question

How do you solve `sqrt(x+3)-sqrt x=sqrt(4x-5)`?

Answer 1

`x=16/11`

Explanation:

This is a tricky equation, so you have first to determine the dominion of it:

`x+3>=0 and x>0 and 4x-5>=0`

`x>=-3 and x>0 and x>=5/4 => x>=5/4`

The standard way to solve this type of equations is to square the parcels, admiting that:

`color(red)(if a=b => a^2=b^2)`

However this brings false solutions, because

`color(red)(if a=-b => a^2=b^2)`

So we have to check the solutions after we obtain the results.

So now let's start:

`sqrt(x+3)-sqrt(x)=sqrt(4x-5)`

`(sqrt(x+3)-sqrt(x))^2=(sqrt(4x-5))^2`

`x+3-2sqrt((x+3)x)+x=4x-5`

Now, you continue to have a "sqrt" in the equation, so you have to square it again. Rearrange the equation in order to isolate the root:

`2sqrt(x^2+3x)=4x-5-x-3-x`

`2sqrt(x^2+3x)=2x-8`

`sqrt(x^2+3x)=x-4`

squaring:

`x^2+3x=x^2-8x+16`

Which gives:

`x=16/11`

First `16/11>5/4?`(the dominion determined above)

Put them in the same denominator:

`(16/11)xx(4/4)>(5/4)xx(11/11) ?`

`64/44>55/44, true`

Now, is the solution true?

`sqrt(16/11+3)-sqrt(16/11)=sqrt(4xx16/11-5)`

`sqrt(49/11)-sqrt(16/11)=sqrt(9/11)`

`(sqrt(49)-sqrt(16))/sqrt(11)=sqrt(9/11)`

`(7-4)/sqrt(11)=3/sqrt(11), true`

Answer 2

`x=16/11`

Explanation:

`1`. When dealing with radicals, try to eliminate them first. Thus, start by squaring both sides of the equation.

`sqrt(x+3)-sqrt(x)=sqrt(4x-5)`

`(sqrt(x+3)-sqrt(x))^2=(sqrt(4x-5))^2`

`2`. Simplify.

`(sqrt(x+3)-sqrt(x))(sqrt(x+3)-sqrt(x))=4x-5`

`x+3-sqrt(x(x+3))-sqrt(x(x+3))+x=4x-5`

`2x+3-sqrt(x^2+3x)-sqrt(x^2+3x)=4x-5`

`-2sqrt(x^2+3x)=2x-8`

`sqrt(x^2+3x)=-1/2(2x-8)`

`sqrt(x^2+3x)=-x+4`

`3`. Since the left side contains a radical, square the whole equation again.

`(sqrt(x^2+3x))^2=(-x+4)^2`

`4`. Simplify.

`(sqrt(x^2+3x))(sqrt(x^2+3x))=(-x+4)(-x+4)`

`x^2+3x=x^2-4x-4x+16`

`color(red)cancelcolor(black)(x^2)+3x=color(red)cancelcolor(black)(x^2)-8x+16`

`3x=-8x+16`

`5`. Solve for `x`.

`11x=16`

`color(green)(x=16/11)`

`:.`, `x` is `16/11`.