Question

How do you solve `sqrt(x+15) - sqrt(2x+7)=1`?

Answer 1

`x = 1`

Explanation:

Even before doing any calculations, you know that any possible solution to this equation must satisfy two conditions

• `x+15>=0 implies x>= - 15`
• `2x+7>=0 implies x >= -7/2`
• `sqrt(x+15)>sqrt(2x+7)`

The third condition implies that `x+15>2x+7 implies x<8`.

Combining these conditions will result in `x in [-7/2, 8)`.

These condtions are required because you can't take the square root of a negative number if you're working with real numbers.

In other words, you can only take the square root of positive real numbers; likewise, taking the square root of a positive number will always result in a positive number.

Start by squaring both sides of the equation; this will reduce the number of radical terms from two to one

`(sqrt(x+15) - sqrt(2x+7))^2 = 1^2`

`(sqrt(x+15))^2 - 2sqrt((x+15)(2x+7)) + (sqrt(2x+7))^2 = 1`

`x + 15 - 2sqrt((x+15)(2x+7)) + 2x + 7 = 1`

Isolate the remaining radical term on one side of the equation

`-2sqrt((x+15)(2x+7)) = 1 - 3x - 22`

`2sqrt((x+15)(2x+7)) = 3x + 21`

Once again, square both sides of the equation to get rid of the radical lterm

`(2sqrt((x+15)(2x+7)))^2 = (3x+21)^2`

`4(x+15)(2x+7) = 9x^2 + 126x + 441`

`8x^2 + 148x + 420 = 9x^2 +126x + 441`

Rearrange this equation to classic quadratic form

`x^2 - 22x + 21 = 0`

You can use the quadratic equation to find the two roots of this quadratic equation

`x_(1,2) = (-(-22) +- sqrt((-22)^2 - 4 * 1 * 21))/(2 * 1)`

`x_(1,2) = (22 +- sqrt(400))/2 = (22 +- 20)/2 = {(x_1 = (22 + 20)/2 = 21), (x_2 = (22-20)/2 = 1) :}`

Notice that `x_1` does not satisfy the condition `x in [-7/2, 8)`, which means that the only valid solution to the original equation is `x_2=1`.

More specifically,

`sqrt(21 + 15) - sqrt(2 * 21 + 7) = 1`

`6 - 7 color(red)(!=) 1 -> x_1` is an extraneous solution

`sqrt(1 + 15) - sqrt(2 * 1 + 7) = 1`

`4 - 3 = 1 color(green)(sqrt())`