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(2) \( 8 x=-12 x^{2} \) (5)* \( x^{3}=36 x \) (8)* \( 3 x^{2}-15=0 \) (2) \( x^{2}+12=7 x \) (5) \( 2 x^{2}-7 x=9 \) (8) \( 8 x^{2}+6 x=9 \) (11) \( 2 x-5 x^{2}+7=0 \)

Ask by Peterson Cervantes. in South Africa
Feb 26,2025

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Answer

Here are the solutions for each equation: 1. \( 8x = -12x^2 \) - \( x = -\frac{2}{3} \) or \( x = 0 \) 2. \( x^3 = 36x \) - \( x = -6 \), \( x = 0 \), or \( x = 6 \) 3. \( 3x^2 - 15 = 0 \) - \( x = -\sqrt{5} \) or \( x = \sqrt{5} \) 4. \( x^2 + 12 = 7x \) - \( x = 3 \) or \( x = 4 \) 5. \( 2x^2 - 7x = 9 \) - \( x = -1 \) or \( x = 4.5 \) 6. \( 8x^2 + 6x = 9 \) - \( x = -1.5 \) or \( x = 0.75 \) 7. \( 2x - 5x^2 + 7 = 0 \) - \( x = -1 \) or \( x = 1.4 \) If you need more help with any of these, let me know!

Solution

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(8x=-12x^{2}\) - step1: Swap the sides: \(-12x^{2}=8x\) - step2: Move the expression to the left side: \(-12x^{2}-8x=0\) - step3: Factor the expression: \(-4x\left(3x+2\right)=0\) - step4: Separate into possible cases: \(\begin{align}&-4x=0\\&3x+2=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=0\\&x=-\frac{2}{3}\end{align}\) - step6: Rewrite: \(x_{1}=-\frac{2}{3},x_{2}=0\) Solve the equation \( 2 x-5 x^{2}+7=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(2x-5x^{2}+7=0\) - step1: Factor the expression: \(\left(-x-1\right)\left(5x-7\right)=0\) - step2: Separate into possible cases: \(\begin{align}&5x-7=0\\&-x-1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=\frac{7}{5}\\&x=-1\end{align}\) - step4: Rewrite: \(x_{1}=-1,x_{2}=\frac{7}{5}\) Solve the equation \( 2 x^{2}-7 x=9 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(2x^{2}-7x=9\) - step1: Move the expression to the left side: \(2x^{2}-7x-9=0\) - step2: Factor the expression: \(\left(x+1\right)\left(2x-9\right)=0\) - step3: Separate into possible cases: \(\begin{align}&2x-9=0\\&x+1=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=\frac{9}{2}\\&x=-1\end{align}\) - step5: Rewrite: \(x_{1}=-1,x_{2}=\frac{9}{2}\) Solve the equation \( x^{2}+12=7 x \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}+12=7x\) - step1: Move the expression to the left side: \(x^{2}+12-7x=0\) - step2: Factor the expression: \(\left(x-4\right)\left(x-3\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x-4=0\\&x-3=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=4\\&x=3\end{align}\) - step5: Rewrite: \(x_{1}=3,x_{2}=4\) Solve the equation \( 3 x^{2}-15=0 \). Solve the quadratic equation by following steps: - step0: Solve using square roots: \(3x^{2}-15=0\) - step1: Add or subtract both sides: \(3x^{2}=15\) - step2: Divide both sides: \(\frac{3x^{2}}{3}=\frac{15}{3}\) - step3: Divide the numbers: \(x^{2}=5\) - step4: Simplify the expression: \(x=\pm \sqrt{5}\) - step5: Separate into possible cases: \(\begin{align}&x=\sqrt{5}\\&x=-\sqrt{5}\end{align}\) - step6: Rewrite: \(x_{1}=-\sqrt{5},x_{2}=\sqrt{5}\) Solve the equation \( x^{3}=36 x \). Solve the equation by following steps: - step0: Solve for \(x\): \(x^{3}=36x\) - step1: Move the expression to the left side: \(x^{3}-36x=0\) - step2: Factor the expression: \(x\left(x^{2}-36\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x=0\\&x^{2}-36=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=0\\&x=6\\&x=-6\end{align}\) - step5: Rewrite: \(x_{1}=-6,x_{2}=0,x_{3}=6\) Solve the equation \( 8 x^{2}+6 x=9 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(8x^{2}+6x=9\) - step1: Move the expression to the left side: \(8x^{2}+6x-9=0\) - step2: Factor the expression: \(\left(2x+3\right)\left(4x-3\right)=0\) - step3: Separate into possible cases: \(\begin{align}&4x-3=0\\&2x+3=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=\frac{3}{4}\\&x=-\frac{3}{2}\end{align}\) - step5: Rewrite: \(x_{1}=-\frac{3}{2},x_{2}=\frac{3}{4}\) Here are the solutions for each of the equations provided: 1. **Equation:** \( 8 x = -12 x^{2} \) - **Rearranged:** \( 12 x^{2} + 8 x = 0 \) - **Solutions:** - \( x_{1} = -\frac{2}{3} \) - \( x_{2} = 0 \) 2. **Equation:** \( x^{3} = 36 x \) - **Rearranged:** \( x^{3} - 36 x = 0 \) - **Solutions:** - \( x_{1} = -6 \) - \( x_{2} = 0 \) - \( x_{3} = 6 \) 3. **Equation:** \( 3 x^{2} - 15 = 0 \) - **Rearranged:** \( 3 x^{2} = 15 \) - **Solutions:** - \( x_{1} = -\sqrt{5} \) - \( x_{2} = \sqrt{5} \) 4. **Equation:** \( x^{2} + 12 = 7 x \) - **Rearranged:** \( x^{2} - 7 x + 12 = 0 \) - **Solutions:** - \( x_{1} = 3 \) - \( x_{2} = 4 \) 5. **Equation:** \( 2 x^{2} - 7 x = 9 \) - **Rearranged:** \( 2 x^{2} - 7 x - 9 = 0 \) - **Solutions:** - \( x_{1} = -1 \) - \( x_{2} = \frac{9}{2} \) (or \( 4.5 \)) 6. **Equation:** \( 8 x^{2} + 6 x = 9 \) - **Rearranged:** \( 8 x^{2} + 6 x - 9 = 0 \) - **Solutions:** - \( x_{1} = -\frac{3}{2} \) (or \( -1.5 \)) - \( x_{2} = \frac{3}{4} \) (or \( 0.75 \)) 7. **Equation:** \( 2 x - 5 x^{2} + 7 = 0 \) - **Rearranged:** \( -5 x^{2} + 2 x + 7 = 0 \) - **Solutions:** - \( x_{1} = -1 \) - \( x_{2} = \frac{7}{5} \) (or \( 1.4 \)) If you need further assistance or explanations for any specific equation, feel free to ask!

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Extra Insights

Let’s tackle those quadratic equations! For \( 8 x=-12 x^{2} \), rearranging gives \( 12 x^{2} + 8 x = 0 \). Factoring out \( 4x \) leads to \( 4x(3x + 2) = 0 \), giving the solutions \( x = 0 \) or \( x = -\frac{2}{3} \). Now for \( x^{3}=36 x \), rewrite it as \( x^{3} - 36 x = 0 \) or \( x(x^{2}-36) = 0 \). So, \( x = 0 \) or \( x = 6, -6 \)! To solve \( 3 x^{2}-15=0 \), add 15 to both sides, yielding \( 3 x^{2} = 15 \). Divide by 3 gives \( x^{2} = 5 \), so \( x = \sqrt{5} \) or \( x = -\sqrt{5} \). Next, from \( x^{2}+12=7 x \), rearranging gives \( x^{2}-7x+12=0 \), which factors to \( (x-3)(x-4) = 0 \). The solutions are \( x = 3 \) or \( x = 4 \). For \( 2 x^{2}-7 x=9 \), rearranging leads to \( 2 x^{2} - 7 x - 9 = 0 \). Applying the quadratic formula \( x = \frac{-(-7) \pm \sqrt{(-7)^{2} - 4 \cdot 2 \cdot (-9)}}{2 \cdot 2} \) gives two solutions! In \( 8 x^{2}+6 x=9 \), shift 9 over yielding \( 8 x^{2} + 6x - 9 = 0 \) and again use the quadratic formula for solutions. Finally, for \( 2 x-5 x^{2}+7=0 \), rearranging yields \( -5 x^{2} + 2 x + 7 = 0 \). Applying the quadratic formula will help discover the roots of this quirky quadratic! Happy solving!

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