\displaystyle \textbf{Question 1: }\text{Solve the following quadratic equations by factorization method:}
\displaystyle \text{i) }x^2+10ix-21=0\qquad\text{ii) }x^2+(1-2i)x-2i=0
\displaystyle \text{iii) }x^2-(2\sqrt{3}+3i)x+6\sqrt{3}i=0
\displaystyle \text{iv) }6x^2-17ix-12=0
\displaystyle \text{Answer:}
\displaystyle \text{i) }x^2+10ix-21=0
\displaystyle \Rightarrow x^2+7ix+3ix-21=0
\displaystyle \Rightarrow x(x+7i)+3i(x+7i)=0
\displaystyle \Rightarrow (x+7i)(x+3i)=0
\displaystyle \Rightarrow x=-7i\text{ or }x=-3i
\displaystyle \therefore \text{The roots of the given quadratic equation are }-7i\text{ and }-3i.
\displaystyle \\

\displaystyle \text{ii) }x^2+(1-2i)x-2i=0
\displaystyle \Rightarrow x^2+x-2ix-2i=0
\displaystyle \Rightarrow x(x+1)-2i(x+1)=0
\displaystyle \Rightarrow (x+1)(x-2i)=0
\displaystyle \Rightarrow x=-1\text{ or }x=2i
\displaystyle \therefore \text{The roots of the given quadratic equation are }-1\text{ and }2i.
\displaystyle \\

\displaystyle \text{iii) }x^2-(2\sqrt{3}+3i)x+6\sqrt{3}i=0
\displaystyle \Rightarrow x^2-2\sqrt{3}x-3ix+6\sqrt{3}i=0
\displaystyle \Rightarrow x(x-2\sqrt{3})-3i(x-2\sqrt{3})=0
\displaystyle \Rightarrow (x-2\sqrt{3})(x-3i)=0
\displaystyle \Rightarrow x=2\sqrt{3}\text{ or }x=3i
\displaystyle \therefore \text{The roots of the given quadratic equation are }2\sqrt{3}\text{ and }3i.
\displaystyle \\

\displaystyle \text{iv) }6x^2-17ix-12=0
\displaystyle \Rightarrow 6x^2-9ix-8ix-12=0
\displaystyle \Rightarrow 3x(2x-3i)-4i(2x-3i)=0
\displaystyle \Rightarrow (2x-3i)(3x-4i)=0
\displaystyle \Rightarrow x=\frac{3}{2}i\text{ or }x=\frac{4}{3}i
\displaystyle \therefore \text{The roots of the given quadratic equation are }\frac{3}{2}i\text{ and }\frac{4}{3}i.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following quadratic equations:}
\displaystyle \text{i) }x^2-(3\sqrt{2}+2i)x+6\sqrt{2}i=0
\displaystyle \text{ii) }x^2-(5-i)x+(18+i)=0
\displaystyle \text{iii) }(2+i)x^2-(5-i)x+2(1-i)=0
\displaystyle \text{iv) }x^2-(2+i)x-(1-7i)=0
\displaystyle \text{v) }ix^2-4x-4i=0\qquad\text{vi) }x^2+4ix-4=0
\displaystyle \text{vii) }2x^2+\sqrt{15}ix-i=0
\displaystyle \text{viii) }x^2-x+(1+i)=0\qquad\text{ix) }ix^2-x+12i=0
\displaystyle \text{x) }x^2-(3\sqrt{2}-2i)x-\sqrt{2}i=0
\displaystyle \text{xi) }x^2-(\sqrt{2}+i)x+\sqrt{2}i=0
\displaystyle \text{xii) }2x^2-(3+7i)x+(9i-3)=0
\displaystyle \text{Answer:}

\displaystyle \text{i) }x^2-(3\sqrt{2}+2i)x+6\sqrt{2}i=0
\displaystyle \Rightarrow x^2-3\sqrt{2}x-2ix+6\sqrt{2}i=0
\displaystyle \Rightarrow x(x-3\sqrt{2})-2i(x-3\sqrt{2})=0
\displaystyle \Rightarrow (x-3\sqrt{2})(x-2i)=0
\displaystyle \Rightarrow x=3\sqrt{2}\text{ or }x=2i
\displaystyle \therefore \text{The roots of the given quadratic equation are }3\sqrt{2}\text{ and }2i.
\displaystyle \\

\displaystyle \text{ii) }x^2-(5-i)x+(18+i)=0
\displaystyle \Rightarrow x^2-(3-4i)x-(2+3i)x+(18+i)=0
\displaystyle \Rightarrow x[x-(3-4i)]-(2+3i)[x-(3-4i)]=0
\displaystyle \Rightarrow [x-(3-4i)][x-(2+3i)]=0
\displaystyle \Rightarrow x=3-4i\text{ or }x=2+3i
\displaystyle \therefore \text{The roots of the given quadratic equation are }3-4i\text{ and }2+3i.
\displaystyle \\

\displaystyle \text{iii) }(2+i)x^2-(5-i)x+2(1-i)=0
\displaystyle \Rightarrow (2+i)x^2-(3-i)x-2x+2(1-i)=0
\displaystyle \Rightarrow (2+i)x[x-(1-i)]-2[x-(1-i)]=0
\displaystyle \Rightarrow [x-(1-i)][(2+i)x-2]=0
\displaystyle \Rightarrow x=1-i\text{ or }x=\frac{2}{2+i}
\displaystyle \Rightarrow x=1-i\text{ or }x=\frac{2(2-i)}{(2+i)(2-i)}
\displaystyle \Rightarrow x=1-i\text{ or }x=\frac{4-2i}{5}
\displaystyle \therefore \text{The roots of the given quadratic equation are }1-i\text{ and }\frac{4-2i}{5}.
\displaystyle \\

\displaystyle \text{iv) }x^2-(2+i)x-(1-7i)=0
\displaystyle \Rightarrow x^2-(3-i)x+(1-2i)x-(1-7i)=0
\displaystyle \Rightarrow x[x-(3-i)]+(1-2i)[x-(3-i)]=0
\displaystyle \Rightarrow [x-(3-i)][x+(1-2i)]=0
\displaystyle \Rightarrow x=3-i\text{ or }x=-1+2i
\displaystyle \therefore \text{The roots of the given quadratic equation are }3-i\text{ and }-1+2i.
\displaystyle \\

\displaystyle \text{v) }ix^2-4x-4i=0
\displaystyle \Rightarrow i(x^2+4ix-4)=0
\displaystyle \Rightarrow i(x+2i)^2=0
\displaystyle \Rightarrow (x+2i)^2=0
\displaystyle \Rightarrow x=-2i
\displaystyle \therefore \text{The given quadratic equation has equal roots, each equal to }-2i.
\displaystyle \\

\displaystyle \text{vi) }x^2+4ix-4=0
\displaystyle \Rightarrow x^2+2ix+2ix-4=0
\displaystyle \Rightarrow x(x+2i)+2i(x+2i)=0
\displaystyle \Rightarrow (x+2i)^2=0
\displaystyle \Rightarrow x=-2i
\displaystyle \therefore \text{The given quadratic equation has equal roots, each equal to }-2i.
\displaystyle \\

\displaystyle \text{vii) }2x^2+\sqrt{15}ix-i=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=2,\ b=\sqrt{15}i,\ c=-i.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{-\sqrt{15}i\pm\sqrt{(\sqrt{15}i)^2-4\times2\times(-i)}}{4}
\displaystyle =\frac{-\sqrt{15}i\pm\sqrt{-15+8i}}{4}
\displaystyle \because (1+4i)^2=1+8i-16=-15+8i
\displaystyle \therefore \sqrt{-15+8i}=1+4i
\displaystyle \Rightarrow x=\frac{-\sqrt{15}i\pm(1+4i)}{4}
\displaystyle \Rightarrow x=\frac{1+(4-\sqrt{15})i}{4}\text{ or }x=\frac{-1-(4+\sqrt{15})i}{4}
\displaystyle \therefore \text{The roots of the given quadratic equation are }\frac{1+(4-\sqrt{15})i}{4}\text{ and }\frac{-1-(4+\sqrt{15})i}{4}.
\displaystyle \\

\displaystyle \text{viii) }x^2-x+(1+i)=0
\displaystyle \Rightarrow x^2-(1-i)x-ix+(1+i)=0
\displaystyle \Rightarrow x[x-(1-i)]-i[x-(1-i)]=0
\displaystyle \Rightarrow [x-(1-i)](x-i)=0
\displaystyle \Rightarrow x=1-i\text{ or }x=i
\displaystyle \therefore \text{The roots of the given quadratic equation are }1-i\text{ and }i.
\displaystyle \\

\displaystyle \text{ix) }ix^2-x+12i=0
\displaystyle \text{Multiplying both sides by }-i,
\displaystyle x^2+ix+12=0
\displaystyle \Rightarrow x^2-3ix+4ix+12=0
\displaystyle \Rightarrow x(x-3i)+4i(x-3i)=0
\displaystyle \Rightarrow (x-3i)(x+4i)=0
\displaystyle \Rightarrow x=3i\text{ or }x=-4i
\displaystyle \therefore \text{The roots of the given quadratic equation are }3i\text{ and }-4i.
\displaystyle \\

\displaystyle \text{x) }x^2-(3\sqrt{2}-2i)x-\sqrt{2}i=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=1,\ b=-(3\sqrt{2}-2i),\ c=-\sqrt{2}i.
\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
\displaystyle x=\frac{3\sqrt{2}-2i\pm\sqrt{(-3\sqrt{2}+2i)^2+4\sqrt{2}i}}{2}
\displaystyle =\frac{3\sqrt{2}-2i\pm\sqrt{14-8\sqrt{2}i}}{2}
\displaystyle \because \left(4-\sqrt{2}i\right)^2=14-8\sqrt{2}i
\displaystyle \therefore \sqrt{14-8\sqrt{2}i}=4-\sqrt{2}i
\displaystyle \Rightarrow x=\frac{3\sqrt{2}-2i\pm(4-\sqrt{2}i)}{2}
\displaystyle \Rightarrow x=2+\frac{3\sqrt{2}}{2}-\left(1+\frac{\sqrt{2}}{2}\right)i
\displaystyle \text{or }x=-2+\frac{3\sqrt{2}}{2}+\left(\frac{\sqrt{2}}{2}-1\right)i
\displaystyle \therefore \text{The roots of the given quadratic equation are}
\displaystyle 2+\frac{3\sqrt{2}}{2}-\left(1+\frac{\sqrt{2}}{2}\right)i\text{ and}
\displaystyle -2+\frac{3\sqrt{2}}{2}+\left(\frac{\sqrt{2}}{2}-1\right)i.
\displaystyle \\

\displaystyle \text{xi) }x^2-(\sqrt{2}+i)x+\sqrt{2}i=0
\displaystyle \Rightarrow x^2-\sqrt{2}x-ix+\sqrt{2}i=0
\displaystyle \Rightarrow x(x-\sqrt{2})-i(x-\sqrt{2})=0
\displaystyle \Rightarrow (x-\sqrt{2})(x-i)=0
\displaystyle \Rightarrow x=\sqrt{2}\text{ or }x=i
\displaystyle \therefore \text{The roots of the given quadratic equation are }\sqrt{2}\text{ and }i.
\displaystyle \\

\displaystyle \text{xii) }2x^2-(3+7i)x+(9i-3)=0
\displaystyle \Rightarrow 2x^2-6ix-(3+i)x+(9i-3)=0
\displaystyle \Rightarrow 2x(x-3i)-(3+i)(x-3i)=0
\displaystyle \Rightarrow (x-3i)(2x-3-i)=0
\displaystyle \Rightarrow x=3i\text{ or }2x=3+i
\displaystyle \Rightarrow x=3i\text{ or }x=\frac{3+i}{2}
\displaystyle \therefore \text{The roots of the given quadratic equation are }3i\text{ and }\frac{3+i}{2}.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.