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QUESTION I Solve for \( x \) : \( \begin{array}{l}1.1 \quad x(2 x-7)=-3 \\ 1.2 \quad \sqrt{x+5}+x=3 \quad \text { [correct to } 2 \text { decimal places) } \\ 1.3 \quad-x^{2}+16>0 \\ 1.4 \quad \text { Solve for } x \text { and } y \text { simultaneously, } \\ \quad x-3 y=1 \text { and }(2 x+y-1)(x-y+1)=0 \\ 1.5 \quad \text { Prove that: } \\ \sqrt{\sqrt[1]{a b}} \cdot \sqrt{b \sqrt[y]{a}}=a^{\frac{2+s}{s s}} b^{\frac{b+s}{n s}}\end{array} \)

Ask by Barker Stuart. in South Africa
Feb 27,2025

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1.1: \( x = \frac{1}{2} \) or \( x = 3 \) 1.2: \( x \approx 0.63 \) 1.3: \( -4 < x < 4 \) 1.4: \( (x, y) = (-2, -1) \) or \( \left(\frac{4}{7}, -\frac{1}{7}\right) \) 1.5: The proof requires further analysis.

Solución

Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(-x^{2}+16>0\) - step1: Rewrite the expression: \(-x^{2}+16=0\) - step2: Move the constant to the right side: \(-x^{2}=0-16\) - step3: Remove 0: \(-x^{2}=-16\) - step4: Change the signs: \(x^{2}=16\) - step5: Simplify the expression: \(x=\pm \sqrt{16}\) - step6: Simplify: \(x=\pm 4\) - step7: Separate into possible cases: \(\begin{align}&x=4\\&x=-4\end{align}\) - step8: Determine the test intervals: \(\begin{align}&x<-4\\&-44\end{align}\) - step9: Choose a value: \(\begin{align}&x_{1}=-5\\&x_{2}=0\\&x_{3}=5\end{align}\) - step10: Test the chosen value: \(\begin{align}&x<-4\textrm{ }\textrm{is not a solution}\\&-44\textrm{ }\textrm{is not a solution}\end{align}\) - step11: The final solution is \(-4 0 \] Rearranging gives: \[ x^2 < 16 \] Thus: \[ -4 < x < 4 \] ### 1.4 Solve the system of equations: The equations are: 1. \( x - 3y = 1 \) 2. \( (2x + y - 1)(x - y + 1) = 0 \) From the first equation, we can express \( x \) in terms of \( y \): \[ x = 3y + 1 \] Substituting into the second equation gives two cases: 1. \( 2(3y + 1) + y - 1 = 0 \) 2. \( (3y + 1) - y + 1 = 0 \) Solving these leads to: \[ (x, y) = (-2, -1) \cup (x, y) = \left(\frac{4}{7}, -\frac{1}{7}\right) \] ### 1.5 Prove the equality: The expression to prove is: \[ \sqrt{\sqrt[1]{ab}} \cdot \sqrt{b \sqrt[y]{a}} = a^{\frac{2+s}{ss}} b^{\frac{b+s}{ns}} \] This calculation timed out, indicating it may be too complex for the current method. ### Summary of Results: 1.1: \( x_{1} = \frac{1}{2}, x_{2} = 3 \) 1.2: \( x \approx 0.63 \) 1.3: \( -4 < x < 4 \) 1.4: \( (x, y) = (-2, -1) \cup (x, y) = \left(\frac{4}{7}, -\frac{1}{7}\right) \) 1.5: Proving the equality requires further analysis due to timeout. If you need further assistance with the proof or any other part, please let me know!

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Beyond the Answer

First additional information: Understanding how to solve quadratic equations like \( x(2x - 7) = -3 \) can boost your algebra skills! Rearranging to standard form \( 2x^2 - 7x + 3 = 0 \) allows you to apply the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). It's like solving a puzzle where each piece leads closer to the answer! Second additional information: When dealing with inequalities like \( -x^2 + 16 > 0 \), remember that rewriting it as \( x^2 < 16 \) gives \( -4 < x < 4 \). It's essential to analyze the solutions' impact on any real-world scenarios, such as predicting range limits, just like racing cars needing to stay within certain speed bounds for a safe track experience!

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