Pregunta
Matsavana Circuit Grade \( 1 / \) Mathematics Investigation INVESTIGATION \# 2 2.1 State the rules for surds. \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) (5). 2.2 When we simplify surds, we often leave a square-root or cube-root in the denominator. However, the calculator rationalises the answer so that there is no surd in the denominator. With that said, rationalise and solve for \( x \) in the following: a. \( \frac{3}{1-\sqrt{2}} \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) (3). \( \qquad \) (4). c. \( \sqrt{3 x^{2}}-\sqrt{12}=0 \) \[ \sqrt{3 x^{2}}-\sqrt{12}=0 \] d. \( \sqrt{18}-x \sqrt{2}=\sqrt{32} \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) (4). b. \[ \sqrt[3]{4+x}=3 \] \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) (4) Page 3 of 4
Ask by Chan Santiago. in South Africa
Mar 09,2025
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**2.1 Rules for Surds**
1. **Product Rule:**
\(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\).
2. **Quotient Rule:**
\(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\) provided \(b \neq 0\).
3. **Simplification of Powers:**
\(\sqrt{a^2} = |a|\).
4. **Addition/Subtraction:**
Surds can only be combined if they are like terms. For example,
\(3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}\).
5. **Rationalisation of the Denominator:**
Multiply numerator and denominator by the conjugate or an appropriate surd factor to remove a surd from the denominator.
---
**2.2 Rationalise and Solve**
**(a) Expression:**
\(\frac{3}{1-\sqrt{2}}\)
*Solution:*
\[
\frac{3}{1-\sqrt{2}} = -3(1+\sqrt{2})
\]
**(b) Equation:**
\(\sqrt[3]{4+x}=3\)
*Solution:*
\[
x = 23
\]
**(c) Equation:**
\(\sqrt{3x^2} - \sqrt{12} = 0\)
*Solution:*
\[
x = 2 \quad \text{or} \quad x = -2
\]
**(d) Equation:**
\(\sqrt{18} - x\sqrt{2} = \sqrt{32}\)
*Solution:*
\[
x = -1
\]
Solución
**2.1 Rules for Surds**
1. **Product Rule:**
\(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\).
2. **Quotient Rule:**
\(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\) provided \(b \neq 0\).
3. **Simplification of Powers:**
\(\sqrt{a^2} = |a|\).
4. **Addition/Subtraction:**
Surds can only be combined if they are like terms. For example,
\(3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}\).
5. **Rationalisation of the Denominator:**
To remove a surd from the denominator, multiply the numerator and denominator by the conjugate (if the denominator is binomial) or an appropriate surd factor.
---
**2.2 Rationalise and Solve**
**(a) Expression:**
\(\frac{3}{1-\sqrt{2}}\)
*Step 1: Multiply numerator and denominator by the conjugate of the denominator*
Conjugate of \(1-\sqrt{2}\) is \(1+\sqrt{2}\).
\[
\frac{3}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}} = \frac{3(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}
\]
*Step 2: Simplify the denominator using the difference of two squares*
\[
(1-\sqrt{2})(1+\sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1-2=-1.
\]
*Step 3: Write the simplified expression*
\[
\frac{3(1+\sqrt{2})}{-1} = -3(1+\sqrt{2}).
\]
---
**(b) Equation:**
\(\sqrt[3]{4+x}=3\)
*Step 1: Cube both sides to remove the cube root*
\[
\left(\sqrt[3]{4+x}\right)^3 = 3^3 \quad \Rightarrow \quad 4+x=27.
\]
*Step 2: Solve for \(x\)*
\[
x = 27-4 = 23.
\]
---
**(c) Equation:**
\(\sqrt{3x^2} - \sqrt{12} = 0\)
*Step 1: Isolate the surd*
\[
\sqrt{3x^2} = \sqrt{12}.
\]
*Step 2: Recognise that \(\sqrt{3x^2} = \sqrt{3}\,|x|\). Also, simplify \(\sqrt{12}\)*
\[
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}.
\]
Thus,
\[
\sqrt{3}\,|x| = 2\sqrt{3}.
\]
*Step 3: Divide both sides by \(\sqrt{3}\)*
\[
|x| = 2.
\]
*Step 4: Solve the absolute value equation*
\[
x = 2 \quad \text{or} \quad x = -2.
\]
---
**(d) Equation:**
\(\sqrt{18} - x\sqrt{2} = \sqrt{32}\)
*Step 1: Simplify the square roots*
\[
\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}, \quad \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}.
\]
*Step 2: Substitute the simplified surds into the equation*
\[
3\sqrt{2} - x\sqrt{2} = 4\sqrt{2}.
\]
*Step 3: Divide the entire equation by \(\sqrt{2}\) (since \(\sqrt{2} \neq 0\))*
\[
3 - x = 4.
\]
*Step 4: Solve for \(x\)*
\[
-x = 4 - 3 \quad \Rightarrow \quad -x = 1 \quad \Rightarrow \quad x = -1.
\]
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Simplifique esta solución Mind Expander
Surds, or irrational roots, follow a few simple rules: you cannot simplify a surd unless you can express it as a product involving a perfect square or cube, like \(\sqrt{4}\) equals \(2\). Moreover, when adding or subtracting surds, you can only combine like terms, just as you would with regular variables. Additionally, when multiplying surds, you can multiply what's inside the roots, such as \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\). To rationalize the denominator in \(\frac{3}{1 - \sqrt{2}}\), multiply the numerator and denominator by the conjugate, \(1 + \sqrt{2}\). This results in \(\frac{3(1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})}\). Simplifying the denominator gives \(1^2 - (\sqrt{2})^2 = 1 - 2 = -1\), leading to \(-3(1 + \sqrt{2})\). For \( \sqrt{3x^2} - \sqrt{12} = 0\), you isolate \(\sqrt{3x^2} = \sqrt{12}\) and square both sides, resulting in \(3x^2 = 12\) and \(x = \pm 2\).
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