Quadratic equations (common fada) Solve for \( x \) : a. \( x^{2}-11 x=0 \) b. \( 2 x^{2}=3 x \) c. \( 12 x^{2}=10 x \) d. \( 5 x(x-2)+3(x-2)=0 \) e \( (2 x-3)(x+2)-5(x+2)=0 \) f. \( 5(x-7)^{2}-(x-7)=0 \) i \( (x-3)^{2}=x(x-3) \)

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-11x=0\) - step1: Factor the expression: \(x\left(x-11\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-11=0\\&x=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=11\\&x=0\end{align}\) - step4: Rewrite: \(x_{1}=0,x_{2}=11\) Solve the equation \( (x-3)^{2}=x(x-3) \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left(x-3\right)^{2}=x\left(x-3\right)\) - step1: Multiply the terms: \(\left(x-3\right)^{2}=x^{2}-3x\) - step2: Expand the expression: \(x^{2}-6x+9=x^{2}-3x\) - step3: Cancel equal terms: \(-6x+9=-3x\) - step4: Move the variable to the left side: \(-6x+9+3x=0\) - step5: Add the terms: \(-3x+9=0\) - step6: Move the constant to the right side: \(-3x=0-9\) - step7: Remove 0: \(-3x=-9\) - step8: Change the signs: \(3x=9\) - step9: Divide both sides: \(\frac{3x}{3}=\frac{9}{3}\) - step10: Divide the numbers: \(x=3\) Solve the equation \( 12 x^{2}=10 x \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(12x^{2}=10x\) - step1: Move the expression to the left side: \(12x^{2}-10x=0\) - step2: Factor the expression: \(2x\left(6x-5\right)=0\) - step3: Separate into possible cases: \(\begin{align}&6x-5=0\\&2x=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=\frac{5}{6}\\&x=0\end{align}\) - step5: Rewrite: \(x_{1}=0,x_{2}=\frac{5}{6}\) Solve the equation \( 2 x^{2}=3 x \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(2x^{2}=3x\) - step1: Move the expression to the left side: \(2x^{2}-3x=0\) - step2: Factor the expression: \(x\left(2x-3\right)=0\) - step3: Separate into possible cases: \(\begin{align}&2x-3=0\\&x=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=\frac{3}{2}\\&x=0\end{align}\) - step5: Rewrite: \(x_{1}=0,x_{2}=\frac{3}{2}\) Solve the equation \( 5 x(x-2)+3(x-2)=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(5x\left(x-2\right)+3\left(x-2\right)=0\) - step1: Factor the expression: \(\left(x-2\right)\left(5x+3\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-2=0\\&5x+3=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=2\\&x=-\frac{3}{5}\end{align}\) - step4: Rewrite: \(x_{1}=-\frac{3}{5},x_{2}=2\) Solve the equation \( (2 x-3)(x+2)-5(x+2)=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(\left(2x-3\right)\left(x+2\right)-5\left(x+2\right)=0\) - step1: Factor the expression: \(\left(x+2\right)\left(2x-3-5\right)=0\) - step2: Separate into possible cases: \(\begin{align}&2x-3-5=0\\&x+2=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=4\\&x=-2\end{align}\) - step4: Rewrite: \(x_{1}=-2,x_{2}=4\) Solve the equation \( 5(x-7)^{2}-(x-7)=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(5\left(x-7\right)^{2}-\left(x-7\right)=0\) - step1: Remove the parentheses: \(5\left(x-7\right)^{2}-x+7=0\) - step2: Expand the expression: \(5x^{2}-71x+252=0\) - step3: Factor the expression: \(\left(x-7\right)\left(5x-36\right)=0\) - step4: Separate into possible cases: \(\begin{align}&5x-36=0\\&x-7=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=\frac{36}{5}\\&x=7\end{align}\) - step6: Rewrite: \(x_{1}=7,x_{2}=\frac{36}{5}\) Here are the solutions for each of the quadratic equations: a. For the equation \( x^{2}-11 x=0 \): - Factoring gives \( x(x - 11) = 0 \). - Thus, the solutions are: \[ x_{1} = 0, \quad x_{2} = 11 \] b. For the equation \( 2 x^{2}=3 x \): - Rearranging gives \( 2x^{2} - 3x = 0 \). - Factoring gives \( x(2x - 3) = 0 \). - Thus, the solutions are: \[ x_{1} = 0, \quad x_{2} = \frac{3}{2} \quad \text{(or } 1.5\text{)} \] c. For the equation \( 12 x^{2}=10 x \): - Rearranging gives \( 12x^{2} - 10x = 0 \). - Factoring gives \( 2x(6x - 5) = 0 \). - Thus, the solutions are: \[ x_{1} = 0, \quad x_{2} = \frac{5}{6} \quad \text{(or } 0.8333\text{)} \] d. For the equation \( 5 x(x-2)+3(x-2)=0 \): - Factoring gives \( (x - 2)(5x + 3) = 0 \). - Thus, the solutions are: \[ x_{1} = -\frac{3}{5} \quad \text{(or } -0.6\text{)}, \quad x_{2} = 2 \] e. For the equation \( (2 x-3)(x+2)-5(x+2)=0 \): - Factoring gives \( (x + 2)(2x - 8) = 0 \). - Thus, the solutions are: \[ x_{1} = -2, \quad x_{2} = 4 \] f. For the equation \( 5(x-7)^{2}-(x-7)=0 \): - Factoring gives \( (x - 7)(5(x - 7) - 1) = 0 \). - Thus, the solutions are: \[ x_{1} = 7, \quad x_{2} = \frac{36}{5} \quad \text{(or } 7.2\text{)} \] i. For the equation \( (x-3)^{2}=x(x-3) \): - Rearranging gives \( (x - 3)(x - 3 - x) = 0 \). - Thus, the solution is: \[ x = 3 \] These are the solutions for each of the quadratic equations provided.