**Radical** **equations** (also known as **irrational**) are equations in which the unknown value appears under a radical sign.

The method for solving radical equation is raising both sides of the equation to the same power.

If we have the equation $\sqrt{f(x)} = g(x)$, then the condition of that equation is always $f(x) \geq 0$, however, this is not a sufficient condition. Respecting the properties of the square root function (the domain of square root function is $\mathbb{R} ^+ \cup \{0\}$), the second condition is $g(x) \geq 0$. Therefore, we need to ensure that both sides of equation are non-negative.

After squaring we have an equivalent equation:

$$f(x) = [g(x)]^2.$$

Condition $f(x) \geq 0$ is now unnecessary (it is automatically satisfied after squaring); the solutions of the equation will thus satisfy condition $g(x) \geq 0$, so that for these solutions it will be $f(x) = [g(x)]^2$.

Now we have:

$\sqrt{f(x)} = g(x) \Leftrightarrow g(x) \geq 0$ and $f(x) = [g(x)]^2$.

In general, this is valid for the square root of every even number $n$:

$\sqrt[n]{f(x)} = g(x) \Leftrightarrow g(x) \geq 0$ and $f(x) = [g(x)]^{n}$.

For the square root of every odd number $n$ it will be

$\sqrt[n]{f(x)} = g(x) \Leftrightarrow f(x) =[g(x)]^{n} $,

because their domain is a whole set of real numbers.

*Example 1.* Solve the equation:

$\sqrt{2x + 1} = 1$.

*Solution*:

Both sides of the equation are non-negative; we can square the equation:

$\sqrt{2x + 1} = 1 / ^2$

$ 2x + 1 = 1$

$ 2x = 0$

$ x = 0$.

We must now confirm if $ x = 0$ it is the correct solution:

$\sqrt{2\cdot 0 +1} = 1$

$\sqrt{1}=1$

$1=1$.

It follows that $x=0$ is the solution of the given equation.

*Example 2*. Solve the equation:

$\sqrt{2x + 1} = \sqrt{x + 2}$.

*Solution*:

Conditions for this equation are $2x+1 \geq 0$ and $x+2 \geq 0 \Rightarrow x\geq -\frac{1}{2}$ and $x\geq -2$.

It follows that $x$ must be in interval $[- \frac{1}{2}, + \infty \rangle$.

Both sides of the equation are always non-negative, therefore we can square the given equation.

$\sqrt{2x + 1} = \sqrt{x + 2} / ^2$

$ 2x + 1 = x + 2$

$ x = 1$.

We need check that $x=1$ is the solution of the initial equation:

$\sqrt{2\cdot 1 + 1} = \sqrt{1 + 2}$

$\sqrt{3}=\sqrt{3}$.

It follows that $x=1$ is the solution of the initial equation. We can conclude that directly from the condition of the equation, without any further requirement to checking.

*Example 3*. Solve the equation:

$\sqrt{x + 1} = 2x – 3$.

*Solution*:

Now we must be sure that the right side of the equation is non-negative. Therefore $2x-3 \geq 0 \Rightarrow x \geq \frac{3}{2}$ is the condition of this equation.

After squaring the equation, we have:

$\sqrt{x + 1} = 2x – 3 \Leftrightarrow x + 1 = 4x^2 – 12x + 9 \Leftrightarrow 4x^2 – 13x + 8 = 0$.

The solutions for quadratic equation $4x^2 – 13x + 8 = 0$ are:

$ x_1 = \frac{13 + \sqrt{41}}{8}$ and $ x_2 = \frac{13 – \sqrt{41}}{8}$.

The only solution is $x_1$ due to satisfied condition $x \geq \frac{3}{2}$.

*Example 4*. Solve the equation:

$\sqrt{x + 1} = 2 + \sqrt{x + 2}$

*Solution*:

Both sides of the equation are always non-negative, therefore we can square the equation. In this example we need to square the equation twice, as displayed below:

$\sqrt{x + 1} = 2 + \sqrt{x + 2} /^2$

$ x + 1 = 4 + 4 \sqrt{x + 2} + x + 2$

$ 4 \sqrt{x + 2} = – 5$

$\sqrt{x + 2} = – \frac{5}{4} /^2$

$ x + 2 = \frac{25}{16}$

$ x = – \frac{7}{16}$.

$ x = – \frac{7}{16}$ is not the solution of the initial equation, because $x \notin [-1, + \infty \rangle$, which is the condition of the equation (check it!).

Note: as we observed through the steps of solving of the equation, that this equation does not have solutions before the second squaring, because the square root cannot be negative.

*Example 5.* Solve the equation.

$\sqrt{2 \sqrt{x + 1}} = 2$.

*Solution*:

Both sides of the equation are non-negative, therefore we can square the equation:

$\sqrt{2 \sqrt{x + 1}} = 2 /^2$

$ 2 \sqrt{x + 1} = 4 /: 2$

$\sqrt{x + 1} = 2 /^2$

$ x + 1 = 4$

$ x = 3$.

Let’s check that $ x = 3$ satisfies the initial equation:

$\sqrt{2 \sqrt{3+1}}=2$

$\sqrt{2 \sqrt{4}}=2$

$\sqrt{2\cdot 2}=2$

$\sqrt{4}=2$

$2=2$.

It follows that $ x = 3$ is the solution of the given equation.

## Radical equations worksheets

**Solve radical equations** (370.6 KiB, 531 hits)