Question

How do you solve `sqrt(x+3)-sqrt(x-1)=1`?

Answer 1

the answer is `x=13/4`

First, let's call `a=sqrt(x+3)` and `b=sqrt(x-1)`

Now use the rule `(a+b)(a-b)=a^2-b^2`
so `4=a^2-b^2=a+b`

So we have `a+b=4` and `a-b=1`

We sum and subtract respectively the two equations and we have

`2a=5`
`2b=3`

So `sqrt(x+3)=5/2 => x = +-25/4 - 3 =>x \in {13/4,-37/4}`
And `sqrt(x-1)=3/2 => x= +- 9/4 + 1 => x \in {13/4,-5/4}`

so `x=13/4`

Answer 2

Firstly know that:

`sqrt(a)sqrt(b)=sqrt(ab)`

And also that:

`sqrt(q)sqrt(q)=q`

Knowing this, let's find the value of x...

`sqrt(x+3)-sqrt(x-1)=1`

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

`(sqrt(x+3)-sqrt(x-1))(sqrt(x+3)-sqrt(x-1))=1`

`x+3-sqrt(x+3)sqrt(x-1)-sqrt(x+3)sqrt(x-1)+(x-1)=1`

`x+3-2sqrt(x+3)sqrt(x-1)+x-1=1`

`2x+2-2sqrt(x+3)sqrt(x-1)=1`

`2(x+1-sqrt(x+3)sqrt(x-1))=1`

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

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

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

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

`x^2+1/2x+1/2x+1/4=x^2-x+3x-3`

`x^2+x+1/4=x^2+2x-3`

`x^2-x^2+1/4+3=2x-x`

`:. x=3+1/4`