\(2^{x+1} -32 = 0\)
1.3 Solving surd equations
Previous
1.2 Rational exponents and surds
|
Next
1.4 Applications of exponentials
|
1.3 Solving surd equations (EMBFB)
We also need to be able to solve equations that involve surds.
Worked example 13: Surd equations
Solve for \(x\): \(5\sqrt[3]{x^4} = \text{405}\)
Write in exponential notation
\begin{align*} 5\left( x^4 \right)^{\frac{1}{3}}&= \text{405} \\ 5x^{\frac{4}{3}}&= \text{405} \end{align*}
Divide both sides of the equation by \(\text{5}\) and simplify
\begin{align*} \frac{5x^{\frac{4}{3}}}{5} &= \frac{\text{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) \\ &= \text{405} \\ &= \text{RHS } \end{align*}
Worked example 14: Surd equations
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*}
Write the final answer
The solution to \(z - 4\sqrt{z} + 3 = 0\) is \(z = 9\) or \(z = 1\).
Worked example 15: Surd equations
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*}
Write the final answer
The solution to \(\sqrt{p-2} - 3 = 0\) is \(p = 11\).
Solving surd equations
Solve for the unknown variable (remember to check that the solution is valid):
\(\text{125} \left ( 3^p \right ) = 27 \left ( 5^p \right )\)
\(2y^{\frac{1}{2}} - 3y^{\frac{1}{4}} + 1 = 0\)
\(t-1 = \sqrt{7-t}\)
\(2z - 7\sqrt{z} + 3 = 0\)
\(x^{\frac{1}{3}}(x^{\frac{1}{3}} + 1) = 6\)
\(2^{4n} - \dfrac{1}{\sqrt[4]{16}} = 0\)
\(\sqrt{31 -10d} = 4 - d\)
\(y - 10\sqrt{y} + 9 = 0\)
\(f = 2 + \sqrt{19 - 2f}\)
Previous
1.2 Rational exponents and surds
|
Table of Contents |
Next
1.4 Applications of exponentials
|