Question

How do you solve `sqrt(x+4) +sqrt(2x-1)=3sqrt(x-1)`?

Answer 1

Explanation:

Before solving, let's determine the restrictions on the variable. A radical is undefined it the number inside is less than 0.

We must therefore set up an inequality and solve:

`sqrt(x + 4) >= 0`

`sqrt(2x - 1) >= 0`

and

`3sqrt(x - 1) >= 0`

Solving each we get that

`x >= -4`

`x >= 1/2`

and

`x >= 1`

We must pick the largest of these, since if we pick one of the smaller ones the equation will become undefined. The largest is `x >=1`; this is our restriction statement. To justify the point I made earlier in this paragraph, try substituting a number smaller than `1` into the equation. For example, `x= -2`

`sqrt(-2 + 4) + sqrt(2 xx -2 - 1) = 3sqrt(-2 - 1)`

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

In this case, two radicals are undefined. Now that we have defined any restrictions on the variable, we can proceed with solving. This can be started by squaring both sides

`(sqrt(x + 4) + sqrt(2x - 1))^2 = (3sqrt(x - 1))^2`

`x + 4 + 2sqrt(x + 4)sqrt(2x - 1) + 2x - 1 = 9(x - 1)`

`3x + 3 + 2sqrt(x + 4)sqrt(2x - 1) = 9x - 9`

`2sqrt(x + 4)sqrt(2x - 1) = 6x - 12`

`sqrt(x + 4)(sqrt(2x - 1)) =( cancel(2)(3x - 6))/cancel(2)`

Since the radicals are both square roots, we can combine them in multiplication.

`sqrt((x + 4)(2x - 1)) = 3x - 6`

`sqrt(2x^2 + 8x - x - 4) = 3x - 6`

`(sqrt(2x^2 + 7x - 4))^2 = (3x - 6)^2`

`2x^2 + 7x - 4 = 9x^2 - 36x + 36`

`0 = 7x^2 - 43x + 40`

`0 = 7x^2 - 8x - 35x + 40`

`0 = 7x(x - 8/7) - 35(x - 8/7)`

`0 = (7x - 35)(x - 8/7)`

`x = 5 and 8/7`

Checking both answers in the original equation, we find that only `x = 5` works.

Therefore, our solution set is `{x = 5, x >= 1}`

Hopefully this helps!