Rewrite the equation so that one square root, such as $\sqrt{x+4}$, is alone on one side of the equation. What can you subtract from both sides to do that?

After isolating, you will have something like $(5 - \sqrt{x-1})^2$. Remember that $(a-b)^2 = a^2 - 2ab + b^2$, not just $a^2 - b^2$.

You should get an equation with just one square root equal to a number. Square again to solve for $x$, and always substitute your result back into the original equation to make sure it really works.

In one expression line, type y = sqrt(x+4) + sqrt(x-1). In another line, type y = 5.

Notice that $x-1$ must be $\ge 0$, so only $x \ge 1$ makes sense. On the graph, you only need to look at the part of the curves where $x$ is at least $1$.

Use the intersection tool (tap on where the two graphs cross or click the point that appears) and read off the $x$-coordinate of the intersection; this $x$-value is the solution to the equation.

Because of the square roots, both expressions under the radicals must be nonnegative:

• $x+4 \ge 0$ and $x-1 \ge 0$, so $x \ge 1$.

Start with the equation

Subtract $\sqrt{x-1}$ from both sides to isolate one square root:

Square both sides of

The left side is

For the right side, use $(a-b)^2 = a^2 - 2ab + b^2$ with $a=5$ and $b=\sqrt{x-1}$:

So the equation becomes

Combine like terms on the right:

Subtract $x$ from both sides; they cancel:

Subtract $24$ from both sides:

Divide both sides by $-10$:

From

square both sides:

Add $1$ to both sides:

Check in the original equation:

so $x=5$ satisfies the equation. Thus, the correct value of $x$ is $5$.