d) \( x^{2}-7 x+6 \) f) \( x^{2}-2 \frac{1}{5} x+\frac{2}{5} \) h) \( 2 x^{2}-11 x+5 \)

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-7x+6=0\) - step1: Factor the expression: \(\left(x-6\right)\left(x-1\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-6=0\\&x-1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=6\\&x=1\end{align}\) - step4: Rewrite: \(x_{1}=1,x_{2}=6\) Solve the equation \( 2 x^{2}-11 x+5=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(2x^{2}-11x+5=0\) - step1: Factor the expression: \(\left(x-5\right)\left(2x-1\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-5=0\\&2x-1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=5\\&x=\frac{1}{2}\end{align}\) - step4: Rewrite: \(x_{1}=\frac{1}{2},x_{2}=5\) Solve the equation \( x^{2}-2 \frac{1}{5} x+\frac{2}{5}=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-2\frac{1}{5}\times x+\frac{2}{5}=0\) - step1: Convert the expressions: \(x^{2}-\frac{11}{5}x+\frac{2}{5}=0\) - step2: Factor the expression: \(\frac{1}{5}\left(x-2\right)\left(5x-1\right)=0\) - step3: Divide the terms: \(\left(x-2\right)\left(5x-1\right)=0\) - step4: Separate into possible cases: \(\begin{align}&x-2=0\\&5x-1=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=2\\&x=\frac{1}{5}\end{align}\) - step6: Rewrite: \(x_{1}=\frac{1}{5},x_{2}=2\) Let's solve each of the quadratic equations step by step. ### d) \( x^{2}-7x+6 = 0 \) To solve this equation, we can factor it: \[ x^{2}-7x+6 = (x-1)(x-6) = 0 \] Setting each factor to zero gives us: 1. \( x - 1 = 0 \) → \( x = 1 \) 2. \( x - 6 = 0 \) → \( x = 6 \) Thus, the solutions are: \[ x_{1} = 1, \quad x_{2} = 6 \] ### f) \( x^{2}-2\frac{1}{5}x+\frac{2}{5} = 0 \) First, let's convert \( 2\frac{1}{5} \) to an improper fraction: \[ 2\frac{1}{5} = \frac{11}{5} \] Now, the equation becomes: \[ x^{2} - \frac{11}{5}x + \frac{2}{5} = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): - \( a = 1 \) - \( b = -\frac{11}{5} \) - \( c = \frac{2}{5} \) Calculating the discriminant: \[ b^2 - 4ac = \left(-\frac{11}{5}\right)^2 - 4 \cdot 1 \cdot \frac{2}{5} = \frac{121}{25} - \frac{8}{5} = \frac{121}{25} - \frac{40}{25} = \frac{81}{25} \] Now, substituting into the quadratic formula: \[ x = \frac{\frac{11}{5} \pm \sqrt{\frac{81}{25}}}{2} = \frac{\frac{11}{5} \pm \frac{9}{5}}{2} \] This gives us two solutions: 1. \( x = \frac{20/5}{2} = 2 \) 2. \( x = \frac{2/5}{2} = \frac{1}{5} \) Thus, the solutions are: \[ x_{1} = \frac{1}{5}, \quad x_{2} = 2 \] ### h) \( 2x^{2}-11x+5 = 0 \) Using the quadratic formula again: - \( a = 2 \) - \( b = -11 \) - \( c = 5 \) Calculating the discriminant: \[ b^2 - 4ac = (-11)^2 - 4 \cdot 2 \cdot 5 = 121 - 40 = 81 \] Now substituting into the quadratic formula: \[ x = \frac{11 \pm \sqrt{81}}{4} = \frac{11 \pm 9}{4} \] This gives us two solutions: 1. \( x = \frac{20}{4} = 5 \) 2. \( x = \frac{2}{4} = \frac{1}{2} \) Thus, the solutions are: \[ x_{1} = \frac{1}{2}, \quad x_{2} = 5 \] ### Summary of Solutions - For \( d) \): \( x_{1} = 1, \quad x_{2} = 6 \) - For \( f) \): \( x_{1} = \frac{1}{5}, \quad x_{2} = 2 \) - For \( h) \): \( x_{1} = \frac{1}{2}, \quad x_{2} = 5 \)