MTH 101 Week 10 Worksheet V1 Find the zeros for each of the following quadratics. If you use the quadratic formula please report \( D \) and write out your formulas. Please round decimal answers to two decimal places. 1. \( x^{2}-12 x+32 \) 2. \( -5 x^{2}-12 x+16 \) The zeros of quaciratic \( x^{2}-12 x+32 \) are \( x=4+x=8 \) Solve the following equations. 3. \( x^{2}+12 x=-11 \) 4. \( 7 x^{2}+14 x=0 \) \[ v=2.22 \text { and } x=-1.05 \text {. } \] 5. \( 6 x^{2}-7 x=14 \) 6. \( 9 x^{2}+4 x+2=3 x^{2}-6 x-9 \)

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}+12x+11=0\) - step1: Factor the expression: \(\left(x+1\right)\left(x+11\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x+1=0\\&x+11=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=-1\\&x=-11\end{align}\) - step4: Rewrite: \(x_{1}=-11,x_{2}=-1\) Solve the equation \( x^{2}-12 x+32=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-12x+32=0\) - step1: Factor the expression: \(\left(x-8\right)\left(x-4\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-8=0\\&x-4=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=8\\&x=4\end{align}\) - step4: Rewrite: \(x_{1}=4,x_{2}=8\) Solve the equation \( 7 x^{2}+14 x=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(7x^{2}+14x=0\) - step1: Factor the expression: \(7x\left(x+2\right)=0\) - step2: Separate into possible cases: \(\begin{align}&7x=0\\&x+2=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=0\\&x=-2\end{align}\) - step4: Rewrite: \(x_{1}=-2,x_{2}=0\) Solve the equation \( 9 x^{2}+4 x+2-3 x^{2}+6 x+9=0 \). Solve the equation(The complex numbers system) by following steps: - step0: Solve using the quadratic formula in the complex numbers system: \(9x^{2}+4x+2-3x^{2}+6x+9=0\) - step1: Calculate: \(6x^{2}+10x+11=0\) - step2: Solve using the quadratic formula: \(x=\frac{-10\pm \sqrt{10^{2}-4\times 6\times 11}}{2\times 6}\) - step3: Simplify the expression: \(x=\frac{-10\pm \sqrt{10^{2}-4\times 6\times 11}}{12}\) - step4: Simplify the expression: \(x=\frac{-10\pm \sqrt{-164}}{12}\) - step5: Simplify the expression: \(x=\frac{-10\pm 2\sqrt{41}\times i}{12}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{-10+2\sqrt{41}\times i}{12}\\&x=\frac{-10-2\sqrt{41}\times i}{12}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=-\frac{5}{6}+\frac{\sqrt{41}}{6}i\\&x=\frac{-10-2\sqrt{41}\times i}{12}\end{align}\) - step8: Simplify the expression: \(\begin{align}&x=-\frac{5}{6}+\frac{\sqrt{41}}{6}i\\&x=-\frac{5}{6}-\frac{\sqrt{41}}{6}i\end{align}\) - step9: Rewrite: \(x_{1}=-\frac{5}{6}-\frac{\sqrt{41}}{6}i,x_{2}=-\frac{5}{6}+\frac{\sqrt{41}}{6}i\) - step10: Remove the complex number(s): \(\textrm{No real solution}\) Solve the equation \( 6 x^{2}-7 x-14=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(6x^{2}-7x-14=0\) - step1: Solve using the quadratic formula: \(x=\frac{7\pm \sqrt{\left(-7\right)^{2}-4\times 6\left(-14\right)}}{2\times 6}\) - step2: Simplify the expression: \(x=\frac{7\pm \sqrt{\left(-7\right)^{2}-4\times 6\left(-14\right)}}{12}\) - step3: Simplify the expression: \(x=\frac{7\pm \sqrt{385}}{12}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{7+\sqrt{385}}{12}\\&x=\frac{7-\sqrt{385}}{12}\end{align}\) - step5: Rewrite: \(x_{1}=\frac{7-\sqrt{385}}{12},x_{2}=\frac{7+\sqrt{385}}{12}\) Solve the equation \( -5 x^{2}-12 x+16=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(-5x^{2}-12x+16=0\) - step1: Multiply both sides: \(5x^{2}+12x-16=0\) - step2: Solve using the quadratic formula: \(x=\frac{-12\pm \sqrt{12^{2}-4\times 5\left(-16\right)}}{2\times 5}\) - step3: Simplify the expression: \(x=\frac{-12\pm \sqrt{12^{2}-4\times 5\left(-16\right)}}{10}\) - step4: Simplify the expression: \(x=\frac{-12\pm \sqrt{464}}{10}\) - step5: Simplify the expression: \(x=\frac{-12\pm 4\sqrt{29}}{10}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{-12+4\sqrt{29}}{10}\\&x=\frac{-12-4\sqrt{29}}{10}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=\frac{-6+2\sqrt{29}}{5}\\&x=\frac{-12-4\sqrt{29}}{10}\end{align}\) - step8: Simplify the expression: \(\begin{align}&x=\frac{-6+2\sqrt{29}}{5}\\&x=-\frac{6+2\sqrt{29}}{5}\end{align}\) - step9: Rewrite: \(x_{1}=-\frac{6+2\sqrt{29}}{5},x_{2}=\frac{-6+2\sqrt{29}}{5}\) Let's solve the problems step by step. ### 1. Find the zeros of \( x^{2}-12 x+32 \) To find the zeros, we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} \] where \( D = b^2 - 4ac \). For the equation \( x^{2}-12 x+32 = 0 \): - \( a = 1 \) - \( b = -12 \) - \( c = 32 \) Calculating \( D \): \[ D = (-12)^2 - 4 \cdot 1 \cdot 32 = 144 - 128 = 16 \] Now, applying the quadratic formula: \[ x = \frac{12 \pm \sqrt{16}}{2 \cdot 1} = \frac{12 \pm 4}{2} \] This gives us: \[ x_1 = \frac{16}{2} = 8, \quad x_2 = \frac{8}{2} = 4 \] **Zeros:** \( x = 4 \) and \( x = 8 \) ### 2. Find the zeros of \( -5 x^{2}-12 x+16 \) Using the quadratic formula again: - \( a = -5 \) - \( b = -12 \) - \( c = 16 \) Calculating \( D \): \[ D = (-12)^2 - 4 \cdot (-5) \cdot 16 = 144 + 320 = 464 \] Now, applying the quadratic formula: \[ x = \frac{-(-12) \pm \sqrt{464}}{2 \cdot (-5)} = \frac{12 \pm \sqrt{464}}{-10} \] Calculating the approximate values: \[ x_1 \approx -3.35, \quad x_2 \approx 0.95 \] **Zeros:** \( x \approx -3.35 \) and \( x \approx 0.95 \) ### 3. Solve \( x^{2}+12 x=-11 \) Rearranging gives: \[ x^{2}+12 x + 11 = 0 \] Using the quadratic formula: - \( a = 1 \) - \( b = 12 \) - \( c = 11 \) Calculating \( D \): \[ D = 12^2 - 4 \cdot 1 \cdot 11 = 144 - 44 = 100 \] Applying the quadratic formula: \[ x = \frac{-12 \pm \sqrt{100}}{2 \cdot 1} = \frac{-12 \pm 10}{2} \] This gives us: \[ x_1 = \frac{-2}{2} = -1, \quad x_2 = \frac{-22}{2} = -11 \] **Zeros:** \( x = -1 \) and \( x = -11 \) ### 4. Solve \( 7 x^{2}+14 x=0 \) Factoring out \( 7x \): \[ 7x(x + 2) = 0 \] This gives: \[ x_1 = 0, \quad x_2 = -2 \] **Zeros:** \( x = 0 \) and \( x = -2 \) ### 5. Solve \( 6 x^{2}-7 x=14 \) Rearranging gives: \[ 6 x^{2}-7 x - 14 = 0 \] Using the quadratic formula: - \( a = 6 \) - \( b = -7 \) - \( c = -14 \) Calculating \( D \): \[ D = (-7)^2 - 4 \cdot 6 \cdot (-14) = 49 + 336 = 385 \] Applying the quadratic formula: \[ x = \frac{7 \pm \sqrt{385}}{12} \] Calculating the approximate values: \[ x_1 \approx -1.05, \quad x_2 \approx 2.22 \] **Zeros:** \( x \approx -1.05 \) and \( x \approx 2.22 \) ### 6. Solve \( 9 x^{2}+4 x+2=3 x^{2}-6 x-9 \) Rearranging gives: \[ 6 x^{2}+10 x+11 = 0 \] Using the quadratic formula: - \( a = 6 \) - \( b = 10 \) - \( c = 11 \) Calculating \( D \): \[ D = 10^2 - 4 \cdot 6 \cdot 11 = 100 - 264 = -164 \] Since \( D < 0 \), there are no real solutions. **Zeros:** No real solution. ### Summary of Results 1. Zeros of \( x^{2}-12 x+32 \): \( x = 4, 8 \) 2. Zeros of \( -5 x^{2}-12 x+16 \): \( x \approx -3.35, 0.95 \) 3. Zeros of \( x^{2}+12 x+11 \): \( x = -1, -11 \) 4. Zeros of \( 7 x^{2}+14 x=0 \): \( x = 0, -2 \) 5. Zeros of \( 6 x^{2}-7 x-14 \): \( x \approx -1.05, 2.22 \) 6. Zeros of \( 9 x^{2}+4 x+2=3 x^{2}-6 x-9 \): No real solution.