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## Solving surd equations

We also need to be able to solve equations that involve surds.

## Example 1: Surd equations

### Question

Solve for $$x$$: $$5\sqrt[3]{x^4} = \textrm{405}$$

Write in exponential notation

\begin{align*} 5\left( x^4 \right)^{\frac{1}{3}}&= \textrm{405} \\ 5x^{\frac{4}{3}}&= \textrm{405} \end{align*}

Divide both sides of the equation by 5 and simplify

\begin{align*} \frac{5x^{\frac{4}{3}}}{5} &= \frac{\textrm{405}}{5} \\ x^{\frac{4}{3}} &= 81 \\ x^{\frac{4}{3}} &= 3^4 \end{align*}

Simplify the exponents

\begin{align} \left( x^{\frac{4}{3}} \right)^{\frac{3}{4}} &= \left( 3^4 \right)^{\frac{3}{4}} \\ x &= 3^3 \\ x &= 27 \end{align}

Check the solution by substituting the answer back into the original equation

\begin{align*} \text{LHS}&= 5\sqrt[3]{x^4} \\ &= 5(27)^{\frac{4}{3}} \\ &= 5(3^3)^{\frac{4}{3}} \\ &= 5(3^4) \\ &= \textrm{405} \\ &= \text{RHS } \end{align*}

## Example 2: Surd equations

### Question

Solve for $$z$$: $$z - 4\sqrt{z} + 3 = 0$$

Factorise

\begin{align} z - 4\sqrt{z} + 3 &= 0 \\ z - 4z^{\frac{1}{2}} + 3 &= 0 \\ (z^{\frac{1}{2}}-3)(z^{\frac{1}{2}}-1) &= 0 \end{align}

Solve for both factors

The zero law states: if $$a \times b = 0$$, then $$a = 0$$ or $$b = 0$$.

$\therefore (z^{\frac{1}{2}}-3) = 0 \text{ or } (z^{\frac{1}{2}}-1) = 0$

Therefore

\begin{align} z^{\frac{1}{2}}-3 &= 0 \\ z^{\frac{1}{2}} &= 3 \\ \left( z^{\frac{1}{2}} \right)^2 &= 3^2 \\ z &= 9 \end{align}

or

\begin{align} z^{\frac{1}{2}}-1 &= 0 \\ z^{\frac{1}{2}} &= 1 \\ \left( z^{\frac{1}{2}} \right)^2 &= 1^2 \\ z &= 1 \end{align}

Check the solution by substituting both answers back into the original equation

If $$z=9$$:

\begin{align} \text{LHS}&= z - 4\sqrt{z} + 3 \\ &= 9 - 4\sqrt{9} + 3 \\ &= 12 - 12 \\ &= 0 \\ &=\text{RHS } \end{align}

If $$z=1$$:

\begin{align} \text{LHS} &= z - 4\sqrt{z} + 3 \\ &= 1 - 4\sqrt{1} + 3 \\ &= 4-4 \\ &= 0 \\ &= \text{RHS } \end{align}

The solution to $$z - 4\sqrt{z} + 3 = 0$$ is $$z = 9$$ or $$z = 1$$.

## Example 3: Surd equations

### Question

Solve for $$p$$: $$\sqrt{p-2} - 3 = 0$$

Write the equation with only the square root on the left hand side

Use the additive inverse to get all other terms on the right hand side and only the square root on the left hand side.

$\sqrt{p-2} = 3$

Square both sides of the equation

\begin{align} \left( \sqrt{p-2} \right)^2 &= 3^2 \\ p-2 &= 9 \\ p &= 11 \end{align}

Check the solution by substituting the answer back into the original equation

If $$p=11$$:

\begin{align} \text{LHS} &=\sqrt{p-2} - 3 \\ &=\sqrt{11-2} - 3 \\ &=\sqrt{9} - 3 \\ &= 3-3 \\ &= 0 \\ &= \text{RHS } \end{align}

The solution to $$\sqrt{p-2} - 3 = 0$$ is $$p = 11$$.

## Exercise 1: Solving surd equations

Solve for the unknown variable (remember to check that the solution is valid):

1. $$2^{x+1} -32 = 0$$

2. $$\textrm{125} \left ( 3^p \right ) = 27 \left ( 5^p \right )$$

3. $$2y^{\frac{1}{2}} - 3y^{\frac{1}{4}} + 1 = 0$$

4. $$t-1 = \sqrt{7-t}$$

5. $$2z - 7\sqrt{z} + 3 = 0$$

6. $$x^{\frac{1}{3}}(x^{\frac{1}{3}} + 1) = 6$$

7. $$2^{4n} - \dfrac{1}{\sqrt[4]{16}} = 0$$

8. $$\sqrt{31 -10d} = 4 - d$$

9. $$y - 10\sqrt{y} + 9 = 0$$

10. $$f = 2 + \sqrt{19 - 2f}$$

1. \begin{aligned} 2^{x+1}-32&=0 \\ 2^{x+1}&=32 \\ 2^{x+1}&=2^5 \\ \therefore x+1&=5 \\ x&=4 \end{aligned}
2. \begin{align*} \textrm{125}\left ( 3^p \right ) &= 27\left ( 5^p \right ) \\ \frac{5^p}{3^p}&=\frac{\textrm{125}}{27} \\ \left ( \frac{5}{3} \right )^p &= \left ( \frac{5}{3} \right )^3 \\ \therefore p&=3 \end{align*}
3. \begin{aligned} 2y^{\frac{1}{2}}-3y^{\frac{1}{4}}+1&=0 \\ \left ( 2y^\frac{1}{4} -1\right )\left ( y^\frac{1}{4} -1\right )&=0 \\ \text{Therefore } 2y^{\frac{1}{4}}-1&=0 \\ y^{\frac{1}{4}}-\frac{1}{2}&=0 \\ y^\frac{1}{4}&=\frac{1}{2} \\ y^\frac{4}{4} &= \left ( \frac{1}{2} \right )^4 \\ \therefore y&=\frac{1}{16} \\ \text{or} \\ y^\frac{1}{4}-1&=0 \\ y^\frac{1}{4}&=1 \\ \therefore y&=1 \end{aligned}
4. \begin{aligned} t-1 &= \sqrt{7-t} \\ \left ( t-1 \right )^2 &= \left ( \sqrt{7-t} \right )^2 \\ t^2 - 2t + 1 &= 7-t \\ t^2 - t - 6 &=0 \\ (t - 3)(t + 2)&= 0 \\ \therefore t =3 &\text{ or } t = - 2 \\ \text{Check RHS for } t = 3: &= \sqrt{7-3} \\ &= \sqrt{4} \\ &= 2 \\ &= \text{LHS} \therefore \text{ valid solution }\\ \text{Check RHS for } t = -2: &= \sqrt{7-(-2)} \\ &= \sqrt{9} \\ &= 3 \\ &\ne \text{LHS} \therefore \text{ not valid solution } \end{aligned}
5. \begin{aligned} 2z-7z^\frac{1}{2}+3&=0 \\ \left ( z^\frac{1}{2}-3 \right )\left ( 2z^\frac{1}{2}-1 \right )&=0 \\ \text{Therefore } z^\frac{1}{2}-3&=0 \\ \left ( z^\frac{1}{2} \right )^2 &=3^2 \\ \therefore z&=9 \\ \text{or}\\ 2z^\frac{1}{2}-1&=0 \\ \left ( z^\frac{1}{2} \right )^2&=\left ( \frac{1}{2} \right )^2 \\ \therefore z&=\frac{1}{4} \end{aligned}
6. \begin{aligned} x^\frac{1}{3}\left ( x^\frac{1}{3}+1 \right )&=6 \\ x^\frac{2}{3}+x^\frac{1}{3} &=6 \\ \left ( x^\frac{1}{3}-2 \right )\left ( x^\frac{1}{3}+3 \right ) &= 0 \\ \text{Therefore } x^\frac{1}{3}-2 &= 0 \\ x^\frac{1}{3} &= 2 \\ \therefore x&=8 \\ \text{or} \\ x^\frac{1}{3}+3 &= 0 \\ x^\frac{1}{3} &=-3 \\ \therefore x &= -27 \end{aligned}
7. \begin{aligned} 2^{4n}-\dfrac{1}{\sqrt[4]{16}}&=0\\ 2^{4n}&=\dfrac{1}{16^\frac{1}{4}} \\ 2^{4n}&=\dfrac{1}{\left (2^4 \right )^\frac{1}{4}} \\ 2^{4n}&=\frac{1}{2} \\ 2^{4n}&=2^{-1} \\ \therefore 4n&=-1 \\ \therefore n &=-\frac{1}{4} \end{aligned}
8. \begin{aligned} \sqrt{31-10d}&=d-4 \\ \left (\sqrt{31-10d} \right )^2&=\left (d-4 \right )^2 \\ 31-10d&=d^2-8d+16 \\ 0&=d^2+2d-15 \\ 0&=\left ( d-3 \right )\left ( d+5 \right ) \\ \text{Therefore } d&= 3 \\ \text{or} \\ d&= -5 \\ \text{Check LHS for } d = 3: &= \sqrt{31 - 30} \\ &= \sqrt{1} \\ &= 1 \\ &= \text{RHS} \therefore \text{ valid solution }\\ \text{Check LHS for } t = -5: &= \sqrt{31-(-50)} \\ &= \sqrt{81} \\ &= 9 \\ &= \text{RHS} \therefore \text{ valid solution } \end{aligned}
9. \begin{aligned} y - 10\sqrt{y} + 9 &= 0 \\ y - 10y^{\frac{1}{2}} + 9 &= 0 \\ \left( y^{\frac{1}{2}} - 1 \right ) \left( y^{\frac{1}{2}} -9 \right ) &= 0 \\ \text{Therefore } y^{\frac{1}{2}} - 1 &= 0 \\ y^{\frac{1}{2}} &= 1 \\ \left( y^{\frac{1}{2}} \right )^2 &= (1)^2 \\ \therefore y &= 1 \\ \text{or} \\ y^{\frac{1}{2}} - 9 & = 0 \\ y^{\frac{1}{2}} &= 9 \\ \left( y^{\frac{1}{2}} \right )^2 &= (9)^2 \\ \therefore y &= 81 \end{aligned}
10. \begin{aligned} f &= 2 + \sqrt{19 - 2f} \\ f - 2 &= \sqrt{19 - 2f} \\ \left( f - 2 \right )^2 &= \left( \sqrt{19 - 2f} \right )^2 \\ f^2 - 4f + 4 &= 19 - 2f \\ f^2 - 2f -15 &= 0 \\ (f - 5)(f + 3) &= 0 \\ \text{Therefore } f - 5 &= 0 \\ \therefore f &= 5 \\ \text{or} \\ f + 3 & = 0 \\ \therefore f &= -3 \\ \text{Check RHS for } f = 5: &= 2 + \sqrt{19 - 10} \\ &= 2 + \sqrt{9} \\ &= 5 \\ &= \text{LHS} \therefore \text{ valid solution }\\ \text{Check RHS for } f = -3: &= 2 + \sqrt{19 + 6}\\ &= 2 + \sqrt{25} \\ &= 7 \\ &\ne \text{LHS} \therefore \text{ not valid solution } \end{aligned}