Question

Solve the following quadratic equations :

(i)$x^2-(3\sqrt{2}+2i)x+6\sqrt{2}i=0$

(ii)$x^2-(5-i)x+(18+i)=0$

(iii)$(2+i)x^2-(5-i)x+2(1-i)=0$

(iv)$x^2-(2+i)x-(1-7i)=0$

(v)$i{x}^2-4x-4i=0$

(vi)$x^2+4ix-4=0$

(vii)$2x^2+\sqrt{15}ix-i=0$

(viii)$x^2-x+(1+i)=0$

(ix)$i{x}^2-x+12i=0$

(x)$x^2-(3\sqrt{2}-2i)x-\sqrt{2}i=0$

(xi)$x^2-(\sqrt{2}+i)x+\sqrt{2}i=0$

(xii)$2x^2-(3+7i)x+(9i-3)=0$

Answer 1

(i)$x^2-(3\sqrt{2}+2i)x+6\sqrt{2}i=0$ $\Rightarrow x^2-3\sqrt{2}x-2ix+\sqrt{2i}=0\Rightarrow x(x-3\sqrt{2})-2i(x-3\sqrt{2})=0\Rightarrow (x-2i)(x-3\sqrt{2})=0\Rightarrow x=2iOr3\sqrt{2}$

(ii)$x^2-(5-i)x+(18+i)=0$ $\Rightarrow x^2-5x-ix+18+i=0\Rightarrow x^2-(3-4i)x-(2+3i)x+(18+i)=0\Rightarrow x(x-(3-4i\left)\right)-(2+3i)(x-(3-4i\left)\right)=0\Rightarrow (x-(2+3i\left)\right)(x-(3-4i\left)\right)=0\Rightarrow x=2+3iOr3-4i$

(iii)$(2+i)x^2-(5-i)x+2(1-i)=0$ $(2+i)x^2-(5-i)x+2(1-i)=0\Rightarrow (2+i)x^2-2x-(3-i)x+2(1-i)=0\Rightarrow x\left[\right(2+i)-2]-(1-i)\left[\right(2+i)x-2]=0\Rightarrow [x-(1-i\left)\right]\left[\right(2+i)x-2]=0either[x-(1-i\left)\right]=0Or\left[\right(2+i)x-2]=0\Rightarrow x=1-iOrx=\frac{2}{2+i}\Rightarrow x=1-iOrx=\frac{2\times 2-i}{(2+i)(2-i)}Orx=\frac{4-2i}{4+1}=\frac{4}{5}-\frac{2}{5}iThus,x=1-i,\frac{4}{5}-\frac{2}{5}i$

(iv)$x^2-(2+i)x-(1-7i)=0$ $\Rightarrow x^2-(2+i)x-(1-7i)=0\Rightarrow x^2-(3-i)x+(1-2i)x-(1-7i)=0\Rightarrow x(x-(3-i\left)\right)+(1-2i)(x-(3-i\left)\right)=0\Rightarrow [x+(1-2i\left)\right][x-(3-i\left)\right]=0\Rightarrow x=-1+2i,3-i$

(v)$i{x}^2-4x-4i=0$ $\Rightarrow i{x}^2+4i^{2}x+4i^3=0[\because i^2=-1]\Rightarrow x^2+4ix+4i^2=0\Rightarrow x^2+2ix+2ix+4i^2=0\Rightarrow x(x+2i)+2i(x+2i)=0\Rightarrow (x+2i)(x+2i)\therefore x=-2i,-2i$

(vi)$x^2+4ix-4=0$ $\Rightarrow x^2+4ix+4i^2=0[\because i^2=-1]\Rightarrow x^2+2ix+2ix+4i^2=0\Rightarrow x(x+2i)+2i(x+2i)=0\Rightarrow (x+2i)(x+2i)=0\Rightarrow x=-2i,-2i$

(vii)$2x^2+\sqrt{15}ix-i=0$ $2x^2+\sqrt{15}ix-i=0\text{Comparing the given equation with the general form}a{x}^2+bx+c=0,wegeta=2,b=\sqrt{15i},c=-i\text{Substituting a and b in,}\alpha =\frac{-b\pm \sqrt{b^2-4ac}}{2a}and\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}\alpha =\frac{-\sqrt{15i}+\sqrt{-15+8i}}{4}and\beta =\frac{-\sqrt{15i}-\sqrt{-15+8i}}{4}Let\sqrt{-15+8i}a+bi\Rightarrow -15+8i=(a+bi{)}^2\Rightarrow -15+8i=a^2-b^2+2abi\Rightarrow a^2-b^2=-15and2abi=8iNow,(a^2-b^2)=(a^2-b^2)+4a^2{b}^2\Rightarrow (a^2+b^2)=(-15{)}^2+64=289\Rightarrow a^2+b^2=17Solving{a}^2-b^2=-15and{a}^2+b^2=17,weget{a}^2=1and{b}^2=16\Rightarrow a=\pm 1andb=\pm 4\Rightarrow a=1,b=4ora=-1,b=-4\therefore \sqrt{-15+8i}=1+4i,-1-4iwhen\sqrt{-15+8i}=1+4i\alpha =\frac{-sqrt15i+1+4i}{4}=\frac{1+(4-\sqrt{15})}{4}iand\beta =\frac{-\sqrt{15i-}(1+4i)}{4}=\frac{-1-(4+\sqrt{15})}{4}iwhen\sqrt{-15+8i}=-1-4i=\alpha =\frac{-sqrt15i-1-4i}{4}=\frac{1+(4-\sqrt{15})}{4}i$

(viii)$x^2-x+(1+i)=0$ $x^2-x+(1+i)=0x^2-ix-(1-i)x+i(1-i)=0(x-i)(x-(1-i\left)\right)=0x=i,1=i$

(ix)$i{x}^2-x+12i=0$ $\text{We will apply discriminant rule,}ona{x}^2+bx+c=0,x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}Now,i{x}^2-x+12i=0x=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(i\left)\right(12i)}}{2i}=\frac{1\pm \sqrt{1+48}}{2i}=\frac{1\pm \sqrt{49}}{2i}=\frac{1\pm 7}{2i}=\frac{8}{2i},\frac{-6}{2i}=\frac{4}{i},\frac{3}{i}=-4i,3i$ (x)$x^2-(3\sqrt{2}-2i)x-\sqrt{2}i=0$ $\text{We will apply discriminant rule,}a{x}^2+bx+c=0,x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}Now,x^2-(3\sqrt{2}-2i)x-\sqrt{2}i=0x=\frac{(3\sqrt{2}-2i)\pm \sqrt{[-(3\sqrt{2}-2i{)}^2]-4(1\left)\right(-\sqrt{2i})}}{2\left(1\right)}=\frac{(3\sqrt{2}-2i)\pm \sqrt{(3\sqrt{2}-2i{)}^2+4\sqrt{2i}}}{2}=\frac{3\sqrt{2}-2i}{2}\pm \frac{4-\sqrt{2i}}{2}$

(xi)$x^2-(\sqrt{2}+i)x+\sqrt{2}i=0$ $x^2\sqrt{2}x-ix+\sqrt{2}i=0x(x-\sqrt{2})-i(x-\sqrt{2})=0(x-i)(x-\sqrt{2})=0x=i,\sqrt{2}$

(xii)$2x^2-(3+7i)x+(9i-3)=0$ $2x^2-3x-7ix+(9i-3)=0(2x-3-i)(x-3i)=0(x-\frac{3+i}{2})(x-3i)=0x=\frac{3+i}{2},3i$