The Correlation Coefficient Notes

<definition term="Radical Equation">An **equation** that **contains** a **variable** **inside** a **radical**.</definition> To solve **radical equations**, we need to **eliminate** the **radical** by **raising** both sides of the equation to the **power** of the **index** of the radical. <callout type="warning">**Squaring** both sides of an equation can **introduce** **extraneous** **solutions**. Therefore, it is **important** to **check** the **solutions** in the **original** **equation**. </callout> ## Solving Radical Equations To solve a **radical equation**: 1. **Isolate** the **radical** on one side of the equation. 2. **Raise** both sides of the equation to the **power** of the **index** of the radical to **eliminate** the radical. 3. **Solve** the resulting equation. 4. **Check** the solutions in the **original** **equation** to **eliminate** any **extraneous** **solutions**. <callout type="example">Solve the equation \\$\sqrt{x} = 5\\$. 1. The radical is already isolated. 2. Square both sides: \\$(\sqrt{x})^2 = 5^2\\$. 3. Simplify: \\$x = 25\\$. 4. Check: \\$\sqrt{25} = 5\\$, so \\$x = 25\\$ is a solution. </callout> <callout type="example">Solve the equation \\$\sqrt{x + 3} = x - 1\\$. 1. The radical is already isolated. 2. Square both sides: \\$(\sqrt{x + 3})^2 = (x - 1)^2\\$. 3. Simplify: \\$x + 3 = x^2 - 2x + 1\\$. 4. Rearrange: \\$x^2 - 3x - 2 = 0\\$. 5. Factor: \\$(x - 2)(x - 1) = 0\\$. 6. Solve: \\$x = 2\\$ or \\$x = 1\\$. 7. Check: 1. For \\$x = 2\\$: \\$\sqrt{2 + 3} = 2 - 1 \Rightarrow \sqrt{5} \neq 1\\$, so \\$x = 2\\$ is not a solution. 2. For \\$x = 1\\$: \\$\sqrt{1 + 3} = 1 - 1 \Rightarrow \sqrt{4} = 0\\$, so \\$x = 1\\$ is a solution. 8. The only solution is \\$x = 1\\$. </callout> <callout type="warning">**Squaring** both sides of an equation can **introduce** **extraneous** **solutions**. Therefore, it is **important** to **check** the **solutions** in the **original** **equation**. </callout> ## Solving Equations with Two Radical Terms When an equation has **two** **radical** **terms**, we need to **eliminate** them **one** at a time. <callout type="example">Solve the equation \\$\sqrt{x + 1} + \sqrt{x - 4} = 5\\$. 1. Isolate one of the radicals: \\$\sqrt{x + 1} = 5 - \sqrt{x - 4}\\$. 2. Square both sides: \\$(\sqrt{x + 1})^2 = (5 - \sqrt{x - 4})^2\\$. 3. Simplify: \\$x + 1 = 25 - 10\sqrt{x - 4} + x - 4\\$. 4. Rearrange: \\$10\sqrt{x - 4} = 20\\$. 5. Divide by 10: \\$\sqrt{x - 4} = 2\\$. 6. Square both sides: \\$(\sqrt{x - 4})^2 = 2^2\\$. 7. Simplify: \\$x - 4 = 4\\$. 8. Solve: \\$x = 8\\$. 9. Check: \\$\sqrt{8 + 1} + \sqrt{8 - 4} = \sqrt{9} + \sqrt{4} = 3 + 2 = 5\\$, so \\$x = 8\\$ is a solution. </callout> <callout type="self_review">Solve the equation \\$\sqrt{x + 2} + \sqrt{x - 3} = 7\\$. </callout> <callout type="tok">How do we know that squaring both sides of an equation is a valid operation? What are the limitations of this method? </callout>