Solving Equations II: Radical Equations

A radical equation is an equation in which a variable appears under a radical sign. It may also have more than one radical. Let’s see some examples of radical equations:

$\sqrt{x-1}=2$

$\sqrt{2x+4}=\sqrt{5}$

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

$\sqrt{x+1}=\sqrt{2x+1}+10$

$\sqrt{10-x}+\sqrt{x-5}=3$

$\sqrt{x^2+3}=3$

$\sqrt[3]{2x-4}=-4$

$\sqrt{\sqrt{x+2}+7}=4$

$\sqrt[3]{x^2-1}=1$

$\sqrt[3]{6x^2+2}-1=x$

If there is only one radical in an equation, it can be solved by isolating the radical and raising both sides to the power necessary to eliminate the radical. Please keep in mind that in solving radical equations, sometimes there are extraneous solutions. Therefore, the answer must be always checked.Let’s see some examples of solving radical equations:

Example 1:

$\sqrt{x-1}=2$

The radical is already isolated. We must raise both sides the the power of $2$ :

$(\sqrt{x-1})^2=2^2$

$x-1=4$

Simplifying:

$x=5$

The answer would be $x=5$ . Now, we must check the solution by substituting $5$ in the original equation for $x$ . If the left side equals the right side after the substitution, we have found the correct answer.

$\sqrt{5-1}=\sqrt{4}=2$

Example 2:

$\sqrt{2x+4}=\sqrt{5}$

Both sides should be raised to the power of $2$ :

$2x+4=5$

Simplify,

$2x=1$

$x=\frac{1}{2}$

Substituting $\frac{1}{2}$ in the original equation for $x$

$\sqrt{2x+4}=\sqrt{2\times \frac{1}{2}+4}=\sqrt{1+4}=\sqrt{5}$

So, the answer $x=\frac{1}{2}$ is correct.

Example 3:

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

Isolating the radical

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

Raising both sides to the power of 2

$-4x+1=(-1-2x)^2$

Simplify

$-4x+1=(-1)^2+(-2x)^2+2\times (-1) \times (-2x)$

$-4x+1=1+4x^2+4x$

$0=4x^2+8x$

$x^2+2x=0$

$x(x+2)=0\to \left\{ \begin{array}{l} x=0 \\ x=-2 \end{array} \right.$

Now, we need to verify both answers. For $x=0$ ,

$\sqrt{-4x+1}+2x=\sqrt{-4 \times 0+1}+2 \times 0=1 \neq -1$

Hence, $x=0$ is not a correct answer. Let’s check $x=-2$ .

$\sqrt{-4x+1}+2x=\sqrt{-4 \times (-2)+1}+2 \times (-2)=\sqrt{9}-4=3-4=1$

Thus, the correct answer is $x=-2$ .

If there are more than one radicals, it may be necessary to isolate radicals in more than one step as shown in the following example.

Example 4:

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

Raising both sides to the power of $2$ :

$(\sqrt{x+1})^2=(\sqrt{2x+3}+1)^2$

$x+1=(\sqrt{2x+3})^2+1^2+2 \times 1 \times \sqrt{2x+3}$

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

Simplify and isolate the radical

$-x-3=2 \sqrt{2x+3}$

Raise both sides to the power of $2$

$(-x-3)^2=2^2 (\sqrt{2x+3})^2$

$x^2 +6x+9=4 (2x+3)$

$x^2 +6x+9=8x+12$

$x^2 -2x-3=0$

This is a quadratic equation and can be solved by the shortened version of the quadratic formula:

$ax^2 + 2kx + c = 0 \to x=\frac{-k \pm \sqrt {k^2 - ac}}{a}$

Therefore,

$x=\frac{1 \pm \sqrt {(-1)^2 + 3}}{1}=1 \pm \sqrt {4}=\left\{ \begin{array}{l} 3 \\ -1 \end{array} \right.$

Now, both answers must be verified. For $x=3$ :

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

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

$2=3-1$

$2=2$

Therefore, $x=3$ is a correct answer.

For $x=-1$ :

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

$\sqrt{-1+1}=\sqrt{2\times (-1)+3}-1$

$0=\sqrt{1}-1$

$0=0$

Hence, $x=-1$ is also a correct answer.

Example 5:

$\sqrt{10-x}+\sqrt{x-5}=3$

We start by isolating one of the radicals

$\sqrt{10-x}=-\sqrt{x-5}+3$

Now, let’s raise both sides by the power of $2$ :

$(\sqrt{10-x})^2=(-\sqrt{x-5}+3)^2$

$10-x=(x-5) +9 -2 \times 3 \times \sqrt{x-5}$

$10-x=x-5 + 9 -6 \sqrt{x-5}$

Simplifying and isolating the second radical:

$6-2x= -6 \sqrt{x-5}$

Raising both sides to the power of $2$ one more time:

$(6-2x)^2= (-6)^2 (\sqrt{x-5})^2$

$36+4x^2-24x= 36 (x-5)$

Simplifying

$4x^2-60x+216= 0$

$x^2-15x+54= 0$

This is a quadratic equation and can be solved by the quadratic formula:

$ax^2 + bx + c = 0 \to x=\frac{-b \pm \sqrt {b^2 - 4ac}}{2a}$

Hence,

$x=\frac{15 \pm \sqrt {15^2 - 4\times 1 \times 54}}{2\times 1}=\frac{15 \pm \sqrt {9}}{2}=\left\{ \begin{array}{l}6 \\9 \end{array} \right.$

We should verify both answers. Substituting $x=6$ in the original equation, we have

$\sqrt{10-x}+\sqrt{x-5}=\sqrt{10-6}+\sqrt{6-5}=2+1 = 3$

Therefore, $x=6$ is a correct answer.

For $x=9$ ,

$\sqrt{10-x}+\sqrt{x-5}=\sqrt{10-9}+\sqrt{9-5}=1+2 = 3$

Consequently, $x=9$ is a correct answer as well.

Example 6:

$\sqrt{x^2+3}=3$

The radical is already isolated. So, we raise both sides by the power of $2$ .

$(\sqrt{x^2+3})^2=3^2$

$x^2+3=9$

Simplify,

$x^2=6 \to x=\pm \sqrt{6}$

$\to \left\{ \begin{array}{l} x=\sqrt{6} \\ x=-\sqrt{6} \end{array} \right.$

Now, we must substitute both answers in the original equation and verify them.

For $x=\sqrt{6}$ ,

$\sqrt{x^2+3}=\sqrt{(\sqrt{6})^2+3}=\sqrt{6+3}=3$

For $x=-\sqrt{6}$ ,

$\sqrt{x^2+3}=\sqrt{(-\sqrt{6})^2+3}=\sqrt{6+3}=3$

There fore, both answers are correct.

Example 7:

$\sqrt[3]{2x-4}=-4$

Here, the radical is isolated and we need to raise both sides by the power of $3$ to eliminate the radical.

$(\sqrt[3]{2x-4})^3=(-4)^3$

$2x-4=-64$

Simplify

$2x=-60$

$x=-30$

To verify, we substitute $-30$ for $x$ in the original equation:

$\sqrt[3]{2x-4}=\sqrt[3]{2\times (-30)-4}=\sqrt[3]{-64}=-4$

Hence, $x=-30$ is a correct answer.

Example 8:

$\sqrt{\sqrt{x+2}+7}=4$

There is a nested radical in this example. Since it is already isolated, we raise both sides to the power of $2$ :

$(\sqrt{\sqrt{x+2}+7})^2=4^2$

$\sqrt{x+2}+7=16$

Now, we isolate the remaining radical:

$\sqrt{x+2}=9$

Raising both sides by the power of $2$

$(\sqrt{x+2})^2=9^2$

Simplify

$x+2=81$

$x=79$

Let’s verify the answer:

$\sqrt{\sqrt{x+2}+7}=\sqrt{\sqrt{79+2}+7}=\sqrt{\sqrt{81}+7}=\sqrt{9+7}=4$

Therefore, $x=79$ is a correct answer.

Example 9:

$\sqrt[3]{x^2-1}=1$

Raise both sides by the third power

$(\sqrt[3]{x^2-1})^3=1^3$

$x^2-1=1$

Simplify

$x^2=2$

$x= \pm \sqrt{2}$

We must verify both answers by substituting in the original equation. For $x=\sqrt{2}$ :

$\sqrt[3]{x^2-1}=\sqrt[3]{(\sqrt{2})^2-1}=\sqrt[3]{2-1}=\sqrt[3]{1}=1.$

Therefore, $x=\sqrt{2}$ is a correct answer. For $x=-\sqrt{2}$ ,

$\sqrt[3]{x^2-1}=\sqrt[3]{(-\sqrt{2})^2-1}=\sqrt[3]{2-1}=\sqrt[3]{1}=1.$

which shows that $x=-\sqrt{2}$ is a correct answer too.

Example 10:

$\sqrt[3]{x^3+2x^2+5}-1=x$

Isolate the radical

$\sqrt[3]{x^3+2x^2+5}=x+1$

Raise both side to the third power

$(\sqrt[3]{x^3+2x^2+5})^3=(x+1)^3$

$x^3+2x^2+5=x^3+3x^2+3x+1$

Simplify

$x^2+3x-4=0$

This is a quadratic equation and can be solved by the quadratic formula:

$ax^2 + bx + c = 0 \to x=\frac{-b \pm \sqrt {b^2 - 4ac}}{2a}$

Therefore,

$x=\frac{-3 \pm \sqrt {3^2 - 4 \times 1 \times (-4)}}{2 \times 1}$

$x=\frac{-3 \pm \sqrt {25}}{2} \to \left\{ \begin{array}{l}x=1 \\ x=-4 \end{array} \right.$

Both answers should be checked by substituting in the original equation.

For $x=1$ :

$\sqrt[3]{x^3+2x^2+5}-1=\sqrt[3]{1^3+2\times 1^2+5}-1=\sqrt[3]{8}-1=2-1=1=x$

For $x=-4$ :

$\sqrt[3]{x^3+2x^2+5}-1=\sqrt[3]{(-4)^3+2\times (-4)^2+5}-1=\sqrt[3]{-64+32+5}-1=-3-1=-4=x$

Hence, both answers are correct.

Comments

2 responses to “Solving Equations II: Radical Equations”

August 22, 2010

emt training

Great information! I’ve been looking for something like this for a while now. Thanks!

December 14, 2010

john madigan

Nice work!

You might also be interested in a free online math solver:
http://www.softmath.com/math-solver/expand-expression.html

Cheers,
John

Solving Equations II: Radical Equations

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