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1. \( x^{2}-5 x+6=0 \) 2. \( x^{3}-3 x^{2}+2 x=0 \) 3 \( x^{2}+7 x=-2 \) 4 \( 7 x^{2}+24 x=15 \) 5. \( (x+2)(x-3) \)

Ask by Christensen Stewart. in South Africa
Feb 25,2025

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Answer

Here are the solutions for each equation: 1. \( x^{2}-5x+6=0 \): \( x=2 \) and \( x=3 \) 2. \( x^{3}-3x^{2}+2x=0 \): \( x=0 \), \( x=1 \), and \( x=2 \) 3. \( x^{2}+7x=-2 \): \( x=-\frac{7+\sqrt{41}}{2} \) and \( x=\frac{-7+\sqrt{41}}{2} \) 4. \( 7x^{2}+24x=15 \): \( x=-\frac{12+\sqrt{249}}{7} \) and \( x=\frac{-12+\sqrt{249}}{7} \) 5. \( (x+2)(x-3)=0 \): \( x=-2 \) and \( x=3 \)

Solution

Solve the equation by following steps: - step0: Solve for \(x\): \(x^{3}-3x^{2}+2x=0\) - step1: Factor the expression: \(x\left(x-2\right)\left(x-1\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x=0\\&x-2=0\\&x-1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=0\\&x=2\\&x=1\end{align}\) - step4: Rewrite: \(x_{1}=0,x_{2}=1,x_{3}=2\) Solve the equation \( x^{2}-5 x+6=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-5x+6=0\) - step1: Factor the expression: \(\left(x-3\right)\left(x-2\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-3=0\\&x-2=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=3\\&x=2\end{align}\) - step4: Rewrite: \(x_{1}=2,x_{2}=3\) Solve the equation \( x^{2}+7 x=-2 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(x^{2}+7x=-2\) - step1: Move the expression to the left side: \(x^{2}+7x+2=0\) - step2: Solve using the quadratic formula: \(x=\frac{-7\pm \sqrt{7^{2}-4\times 2}}{2}\) - step3: Simplify the expression: \(x=\frac{-7\pm \sqrt{41}}{2}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{-7+\sqrt{41}}{2}\\&x=\frac{-7-\sqrt{41}}{2}\end{align}\) - step5: Rewrite the fraction: \(\begin{align}&x=\frac{-7+\sqrt{41}}{2}\\&x=-\frac{7+\sqrt{41}}{2}\end{align}\) - step6: Rewrite: \(x_{1}=-\frac{7+\sqrt{41}}{2},x_{2}=\frac{-7+\sqrt{41}}{2}\) Solve the equation \( 7 x^{2}+24 x=15 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(7x^{2}+24x=15\) - step1: Move the expression to the left side: \(7x^{2}+24x-15=0\) - step2: Solve using the quadratic formula: \(x=\frac{-24\pm \sqrt{24^{2}-4\times 7\left(-15\right)}}{2\times 7}\) - step3: Simplify the expression: \(x=\frac{-24\pm \sqrt{24^{2}-4\times 7\left(-15\right)}}{14}\) - step4: Simplify the expression: \(x=\frac{-24\pm \sqrt{996}}{14}\) - step5: Simplify the expression: \(x=\frac{-24\pm 2\sqrt{249}}{14}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{-24+2\sqrt{249}}{14}\\&x=\frac{-24-2\sqrt{249}}{14}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=\frac{-12+\sqrt{249}}{7}\\&x=\frac{-24-2\sqrt{249}}{14}\end{align}\) - step8: Simplify the expression: \(\begin{align}&x=\frac{-12+\sqrt{249}}{7}\\&x=-\frac{12+\sqrt{249}}{7}\end{align}\) - step9: Rewrite: \(x_{1}=-\frac{12+\sqrt{249}}{7},x_{2}=\frac{-12+\sqrt{249}}{7}\) Solve the equation \( (x+2)(x-3)=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(\left(x+2\right)\left(x-3\right)=0\) - step1: Separate into possible cases: \(\begin{align}&x-3=0\\&x+2=0\end{align}\) - step2: Solve the equation: \(\begin{align}&x=3\\&x=-2\end{align}\) - step3: Rewrite: \(x_{1}=-2,x_{2}=3\) Here are the solutions for each of the equations provided: 1. **Equation:** \( x^{2}-5x+6=0 \) - **Solutions:** \( x_{1}=2, x_{2}=3 \) 2. **Equation:** \( x^{3}-3x^{2}+2x=0 \) - **Solutions:** \( x_{1}=0, x_{2}=1, x_{3}=2 \) 3. **Equation:** \( x^{2}+7x=-2 \) - **Rearranged:** \( x^{2}+7x+2=0 \) - **Solutions:** \( x_{1}=-\frac{7+\sqrt{41}}{2}, x_{2}=\frac{-7+\sqrt{41}}{2} \) 4. **Equation:** \( 7x^{2}+24x=15 \) - **Rearranged:** \( 7x^{2}+24x-15=0 \) - **Solutions:** \( x_{1}=-\frac{12+\sqrt{249}}{7}, x_{2}=\frac{-12+\sqrt{249}}{7} \) 5. **Equation:** \( (x+2)(x-3)=0 \) - **Solutions:** \( x_{1}=-2, x_{2}=3 \) If you need further assistance or explanations for any of these solutions, feel free to ask!

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The Deep Dive

To solve the equation \( x^{2}-5 x+6=0 \), you can factor it into \( (x-2)(x-3)=0 \). This reveals the solutions: \( x=2 \) and \( x=3 \). Always double-check by substituting back into the original equation! For \( x^{3}-3 x^{2}+2 x=0 \), you can factor out \( x \) to obtain \( x(x^{2}-3x+2)=0 \). This factors further into \( x(x-1)(x-2)=0 \), leading to the solutions \( x=0 \), \( x=1 \), and \( x=2 \). Remember, finding a common factor can save time, so keep an eye out for it!

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