\displaystyle \textbf{Question 1: } \text{Solve the quadratic equation by factorization: }x^2+1=0.
\displaystyle \text{Answer:}
\displaystyle x^2+1=0
\displaystyle \Rightarrow x^2-i^2=0
\displaystyle \Rightarrow (x-i)(x+i)=0
\displaystyle \Rightarrow x=i\text{ or }x=-i
\displaystyle \therefore \text{The roots of the equation }x^2+1=0\text{ are }i\text{ and }-i.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{Solve the quadratic equation by factorization: }9x^2+4=0.
\displaystyle \text{Answer:}
\displaystyle 9x^2+4=0
\displaystyle \Rightarrow (3x)^2-(2i)^2=0
\displaystyle \Rightarrow (3x-2i)(3x+2i)=0
\displaystyle \Rightarrow x=\frac{2i}{3}\text{ or }x=-\frac{2i}{3}
\displaystyle \therefore \text{The roots of the equation }9x^2+4=0\text{ are }\frac{2i}{3}\text{ and }-\frac{2i}{3}.
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Solve the quadratic equation by factorization: }x^2+2x+5=0.
\displaystyle \text{Answer:}
\displaystyle x^2+2x+5=0
\displaystyle \Rightarrow x^2+2x+1+4=0
\displaystyle \Rightarrow (x+1)^2-(2i)^2=0
\displaystyle \Rightarrow (x+1-2i)(x+1+2i)=0
\displaystyle \Rightarrow x=-1+2i\text{ or }x=-1-2i
\displaystyle \therefore \text{The roots of the equation }x^2+2x+5=0\text{ are }-1+2i\text{ and }-1-2i.
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{Solve the quadratic equation by factorization: }4x^2-12x+25=0.
\displaystyle \text{Answer:}
\displaystyle 4x^2-12x+25=0
\displaystyle \Rightarrow 4x^2-12x+9+16=0
\displaystyle \Rightarrow (2x-3)^2-(4i)^2=0
\displaystyle \Rightarrow (2x-3-4i)(2x-3+4i)=0
\displaystyle \Rightarrow x=\frac{3+4i}{2}\text{ or }x=\frac{3-4i}{2}
\displaystyle \Rightarrow x=\left(\frac{3}{2}+2i\right)\text{ or }x=\left(\frac{3}{2}-2i\right)
\displaystyle \therefore \text{The roots of the equation }4x^2-12x+25=0\text{ are }\left(\frac{3}{2}+2i\right)\text{ and }\left(\frac{3}{2}-2i\right).
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{Solve the quadratic equation by factorization: }x^2+x+1=0.
\displaystyle \text{Answer:}
\displaystyle x^2+x+1=0
\displaystyle \Rightarrow x^2+x+\frac{1}{4}+\frac{3}{4}=0
\displaystyle \Rightarrow \left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{3}}{2}i\right)^2=0
\displaystyle \Rightarrow \left(x+\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\left(x+\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=0
\displaystyle \Rightarrow x=-\frac{1}{2}+\frac{\sqrt{3}}{2}i\text{ or }x=-\frac{1}{2}-\frac{\sqrt{3}}{2}i
\displaystyle \therefore \text{The roots of the equation }x^2+x+1\text{ are }-\frac{1}{2}+\frac{\sqrt{3}}{2}i\text{ and }-\frac{1}{2}-\frac{\sqrt{3}}{2}i.
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{Solve the quadratic equation by factorization: }4x^2+1=0.
\displaystyle \text{Answer:}
\displaystyle 4x^2+1=0
\displaystyle \Rightarrow (2x)^2-i^2=0
\displaystyle \Rightarrow (2x-i)(2x+i)=0
\displaystyle \Rightarrow x=\frac{i}{2}\text{ or }x=-\frac{i}{2}
\displaystyle \therefore \text{The roots of the equation }4x^2+1=0\text{ are }\frac{i}{2}\text{ and }-\frac{i}{2}.
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{Solve the quadratic equation by factorization: }x^2-4x+7=0.
\displaystyle \text{Answer:}
\displaystyle x^2-4x+7=0
\displaystyle \Rightarrow x^2-4x+4+3=0
\displaystyle \Rightarrow (x-2)^2-(\sqrt{3}i)^2=0
\displaystyle \Rightarrow (x-2-\sqrt{3}i)(x-2+\sqrt{3}i)=0
\displaystyle \Rightarrow x=2+\sqrt{3}i\text{ or }x=2-\sqrt{3}i
\displaystyle \therefore \text{The roots of the equation }x^2-4x+7=0\text{ are }2+\sqrt{3}i\text{ and }2-\sqrt{3}i.
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{Solve the quadratic equation by factorization: }x^2+2x+2=0.
\displaystyle \text{Answer:}
\displaystyle x^2+2x+2=0
\displaystyle \Rightarrow x^2+2x+1+1=0
\displaystyle \Rightarrow (x+1)^2-i^2=0
\displaystyle \Rightarrow (x+1-i)(x+1+i)=0
\displaystyle \Rightarrow x=-1+i\text{ or }x=-1-i
\displaystyle \therefore \text{The roots of the equation }x^2+2x+2=0\text{ are }-1+i\text{ and }-1-i.
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{Solve the quadratic equation by factorization: }5x^2-6x+2=0.
\displaystyle \text{Answer:}
\displaystyle 5x^2-6x+2=0
\displaystyle \Rightarrow x^2-\frac{6}{5}x+\frac{2}{5}=0
\displaystyle \Rightarrow x^2-\frac{6}{5}x+\frac{9}{25}+\frac{1}{25}=0
\displaystyle \Rightarrow \left(x-\frac{3}{5}\right)^2-\left(\frac{i}{5}\right)^2=0
\displaystyle \Rightarrow \left(x-\frac{3}{5}-\frac{i}{5}\right)\left(x-\frac{3}{5}+\frac{i}{5}\right)=0
\displaystyle \Rightarrow x=\frac{3}{5}+\frac{i}{5}\text{ or }x=\frac{3}{5}-\frac{i}{5}
\displaystyle \therefore \text{The roots of the equation }5x^2-6x+2=0\text{ are }\frac{3}{5}+\frac{i}{5}\text{ and }\frac{3}{5}-\frac{i}{5}.
\displaystyle \\

Solve the following quadratics:

\displaystyle \textbf{Question 10: } \text{Solve the quadratic equation }21x^2+9x+1=0.
\displaystyle \text{Answer:}
\displaystyle 21x^2+9x+1=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=21,\ b=9,\ c=1.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-9\pm\sqrt{81-4\times21\times1}}{2\times21}
\displaystyle =\frac{-9\pm\sqrt{-3}}{42}
\displaystyle =\frac{-9\pm\sqrt{3}i}{42}
\displaystyle =-\frac{3}{14}\pm\frac{\sqrt{3}}{42}i
\displaystyle \therefore \text{The roots of the equation are }-\frac{3}{14}\pm\frac{\sqrt{3}}{42}i.
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{Solve the quadratic equation }x^2-x+1=0.
\displaystyle \text{Answer:}
\displaystyle x^2-x+1=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=1,\ b=-1,\ c=1.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{1\pm\sqrt{1-4\times1\times1}}{2}
\displaystyle =\frac{1\pm\sqrt{-3}}{2}
\displaystyle =\frac{1\pm\sqrt{3}i}{2}
\displaystyle =\frac{1}{2}\pm\frac{\sqrt{3}}{2}i
\displaystyle \therefore \text{The roots of the equation are }\frac{1}{2}\pm\frac{\sqrt{3}}{2}i.
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{Solve the quadratic equation }x^2+x+1=0.
\displaystyle \text{Answer:}
\displaystyle x^2+x+1=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=1,\ b=1,\ c=1.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-1\pm\sqrt{1-4\times1\times1}}{2}
\displaystyle =\frac{-1\pm\sqrt{-3}}{2}
\displaystyle =\frac{-1\pm\sqrt{3}i}{2}
\displaystyle =-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i
\displaystyle \therefore \text{The roots of the equation are }-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i.
\displaystyle \\

\displaystyle \textbf{Question 13: } \text{Solve the quadratic equation }17x^2-8x+1=0.
\displaystyle \text{Answer:}
\displaystyle 17x^2-8x+1=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=17,\ b=-8,\ c=1.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{8\pm\sqrt{64-4\times17\times1}}{34}
\displaystyle =\frac{8\pm\sqrt{-4}}{34}
\displaystyle =\frac{8\pm2i}{34}
\displaystyle =\frac{4}{17}\pm\frac{1}{17}i
\displaystyle \therefore \text{The roots of the equation are }\frac{4}{17}\pm\frac{1}{17}i.
\displaystyle \\

\displaystyle \textbf{Question 14: } \text{Solve the quadratic equation }27x^2-10x+1=0.
\displaystyle \text{Answer:}
\displaystyle 27x^2-10x+1=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=27,\ b=-10,\ c=1.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{10\pm\sqrt{100-4\times27\times1}}{54}
\displaystyle =\frac{10\pm\sqrt{-8}}{54}
\displaystyle =\frac{10\pm2\sqrt{2}i}{54}
\displaystyle =\frac{5}{27}\pm\frac{\sqrt{2}}{27}i
\displaystyle \therefore \text{The roots of the equation are }\frac{5}{27}\pm\frac{\sqrt{2}}{27}i.
\displaystyle \\

\displaystyle \textbf{Question 15: } \text{Solve the quadratic equation }17x^2+28x+12=0.
\displaystyle \text{Answer:}
\displaystyle 17x^2+28x+12=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=17,\ b=28,\ c=12.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-28\pm\sqrt{784-4\times17\times12}}{34}
\displaystyle =\frac{-28\pm\sqrt{-32}}{34}
\displaystyle =\frac{-28\pm4\sqrt{2}i}{34}
\displaystyle =-\frac{14}{17}\pm\frac{2\sqrt{2}}{17}i
\displaystyle \therefore \text{The roots of the equation are }-\frac{14}{17}\pm\frac{2\sqrt{2}}{17}i.
\displaystyle \\

\displaystyle \textbf{Question 16: } \text{Solve the quadratic equation }21x^2-28x+10=0.
\displaystyle \text{Answer:}
\displaystyle 21x^2-28x+10=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=21,\ b=-28,\ c=10.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{28\pm\sqrt{784-4\times21\times10}}{42}
\displaystyle =\frac{28\pm\sqrt{-56}}{42}
\displaystyle =\frac{28\pm2\sqrt{14}i}{42}
\displaystyle =\frac{2}{3}\pm\frac{\sqrt{14}}{21}i
\displaystyle \therefore \text{The roots of the equation are }\frac{2}{3}\pm\frac{\sqrt{14}}{21}i.
\displaystyle \\

\displaystyle \textbf{Question 17: } \text{Solve the quadratic equation }8x^2-9x+3=0.
\displaystyle \text{Answer:}
\displaystyle 8x^2-9x+3=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=8,\ b=-9,\ c=3.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{9\pm\sqrt{81-4\times8\times3}}{16}
\displaystyle =\frac{9\pm\sqrt{-15}}{16}
\displaystyle =\frac{9\pm\sqrt{15}i}{16}
\displaystyle =\frac{9}{16}\pm\frac{\sqrt{15}}{16}i
\displaystyle \therefore \text{The roots of the equation are }\frac{9}{16}\pm\frac{\sqrt{15}}{16}i.
\displaystyle \\

\displaystyle \textbf{Question 18: } \text{Solve the quadratic equation }13x^2+7x+1=0.
\displaystyle \text{Answer:}
\displaystyle 13x^2+7x+1=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=13,\ b=7,\ c=1.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-7\pm\sqrt{49-4\times13\times1}}{26}
\displaystyle =\frac{-7\pm\sqrt{-3}}{26}
\displaystyle =\frac{-7\pm\sqrt{3}i}{26}
\displaystyle =-\frac{7}{26}\pm\frac{\sqrt{3}}{26}i
\displaystyle \therefore \text{The roots of the equation are }-\frac{7}{26}\pm\frac{\sqrt{3}}{26}i.
\displaystyle \\

\displaystyle \textbf{Question 19: } \text{Solve the quadratic equation }2x^2+x+1=0.
\displaystyle \text{Answer:}
\displaystyle 2x^2+x+1=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=2,\ b=1,\ c=1.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-1\pm\sqrt{1-4\times2\times1}}{4}
\displaystyle =\frac{-1\pm\sqrt{-7}}{4}
\displaystyle =\frac{-1\pm\sqrt{7}i}{4}
\displaystyle =-\frac{1}{4}\pm\frac{\sqrt{7}}{4}i
\displaystyle \therefore \text{The roots of the equation are }-\frac{1}{4}\pm\frac{\sqrt{7}}{4}i.
\displaystyle \\

\displaystyle \textbf{Question 20: } \text{Solve the quadratic equation }\sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0.
\displaystyle \text{Answer:}
\displaystyle \sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=\sqrt{3},\ b=-\sqrt{2},\ c=3\sqrt{3}.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{\sqrt{2}\pm\sqrt{2-4\times\sqrt{3}\times3\sqrt{3}}}{2\sqrt{3}}
\displaystyle =\frac{\sqrt{2}\pm\sqrt{-34}}{2\sqrt{3}}
\displaystyle =\frac{\sqrt{2}\pm\sqrt{34}\,i}{2\sqrt{3}}
\displaystyle =\frac{1}{\sqrt{6}}\pm\frac{\sqrt{17}}{\sqrt{6}}i
\displaystyle \therefore \text{The roots of the equation are }\frac{1}{\sqrt{6}}\pm\frac{\sqrt{17}}{\sqrt{6}}i.
\displaystyle \\

\displaystyle \textbf{Question 21: } \text{Solve the quadratic equation }\sqrt{2}x^2+x+\sqrt{2}=0.
\displaystyle \text{Answer:}
\displaystyle \sqrt{2}x^2+x+\sqrt{2}=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=\sqrt{2},\ b=1,\ c=\sqrt{2}.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-1\pm\sqrt{1-4\times\sqrt{2}\times\sqrt{2}}}{2\sqrt{2}}
\displaystyle =\frac{-1\pm\sqrt{-7}}{2\sqrt{2}}
\displaystyle =\frac{-1\pm\sqrt{7}i}{2\sqrt{2}}
\displaystyle =-\frac{1}{2\sqrt{2}}\pm\frac{\sqrt{7}}{2\sqrt{2}}i
\displaystyle \therefore \text{The roots of the equation are }-\frac{1}{2\sqrt{2}}\pm\frac{\sqrt{7}}{2\sqrt{2}}i.
\displaystyle \\

\displaystyle \textbf{Question 22: } \text{Solve the quadratic equation }x^2+x+\frac{1}{\sqrt{2}}=0.
\displaystyle \text{Answer:}
\displaystyle x^2+x+\frac{1}{\sqrt{2}}=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=1,\ b=1,\ c=\frac{1}{\sqrt{2}}.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-1\pm\sqrt{1-4\times1\times\frac{1}{\sqrt{2}}}}{2}
\displaystyle =\frac{-1\pm\sqrt{1-2\sqrt{2}}}{2}
\displaystyle =\frac{-1\pm i\sqrt{2\sqrt{2}-1}}{2}
\displaystyle \therefore \text{The roots of the equation are }\frac{-1\pm i\sqrt{2\sqrt{2}-1}}{2}.
\displaystyle \\

\displaystyle \textbf{Question 23: } \text{Solve the quadratic equation }x^2+\frac{x}{\sqrt{2}}+1=0.
\displaystyle \text{Answer:}
\displaystyle x^2+\frac{x}{\sqrt{2}}+1=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=1,\ b=\frac{1}{\sqrt{2}},\ c=1.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-\frac{1}{\sqrt{2}}\pm\sqrt{\frac{1}{2}-4\times1\times1}}{2}
\displaystyle =\frac{-\frac{1}{\sqrt{2}}\pm\sqrt{-\frac{7}{2}}}{2}
\displaystyle =\frac{-1\pm i\sqrt{7}}{2\sqrt{2}}
\displaystyle \therefore \text{The roots of the equation are }\frac{-1\pm i\sqrt{7}}{2\sqrt{2}}.
\displaystyle \\

\displaystyle \textbf{Question 24: } \text{Solve the quadratic equation }\sqrt{5}x^2+x+\sqrt{5}=0.
\displaystyle \text{Answer:}
\displaystyle \sqrt{5}x^2+x+\sqrt{5}=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=\sqrt{5},\ b=1,\ c=\sqrt{5}.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-1\pm\sqrt{1-4\times\sqrt{5}\times\sqrt{5}}}{2\sqrt{5}}
\displaystyle =\frac{-1\pm\sqrt{-19}}{2\sqrt{5}}
\displaystyle =\frac{-1\pm\sqrt{19}\,i}{2\sqrt{5}}
\displaystyle \therefore \text{The roots of the equation are }\frac{-1\pm\sqrt{19}\,i}{2\sqrt{5}}.
\displaystyle \\

\displaystyle \textbf{Question 25: } \text{Solve the quadratic equation }-x^2+x-2=0.
\displaystyle \text{Answer:}
\displaystyle -x^2+x-2=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=-1,\ b=1,\ c=-2.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-1\pm\sqrt{1-4\times(-1)\times(-2)}}{-2}
\displaystyle =\frac{-1\pm\sqrt{-7}}{-2}
\displaystyle =\frac{-1\pm\sqrt{7}\,i}{-2}
\displaystyle =\frac{1\mp\sqrt{7}\,i}{2}
\displaystyle \therefore \text{The roots of the equation are }\frac{1\pm\sqrt{7}\,i}{2}.
\displaystyle \\

\displaystyle \textbf{Question 26: } \text{Solve the quadratic equation }x^2-2x+\frac{3}{2}=0.
\displaystyle \text{Answer:}
\displaystyle x^2-2x+\frac{3}{2}=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=1,\ b=-2,\ c=\frac{3}{2}.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{2\pm\sqrt{(-2)^2-4\times1\times\frac{3}{2}}}{2}
\displaystyle =\frac{2\pm\sqrt{-2}}{2}
\displaystyle =\frac{2\pm\sqrt{2}\,i}{2}
\displaystyle =1\pm\frac{i}{\sqrt{2}}
\displaystyle \therefore \text{The roots of the equation are }1\pm\frac{i}{\sqrt{2}}.
\displaystyle \\

\displaystyle \textbf{Question 27: } \text{Solve the quadratic equation }3x^2-4x+\frac{20}{3}=0.
\displaystyle \text{Answer:}
\displaystyle 3x^2-4x+\frac{20}{3}=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=3,\ b=-4,\ c=\frac{20}{3}.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{4\pm\sqrt{16-4\times3\times\frac{20}{3}}}{6}
\displaystyle =\frac{4\pm\sqrt{-64}}{6}
\displaystyle =\frac{4\pm8i}{6}
\displaystyle =\frac{2\pm4i}{3}
\displaystyle \therefore \text{The roots of the equation are }\frac{2\pm4i}{3}.
\displaystyle \\

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