Solve the following equations for $x, y \in R$ : $( x + iy )(5+6 i )=2+3 i$
Answer
$( x + iy )(5+6 i )=2+3 i$
$\therefore \quad x+ i y=\frac{2+3 i }{5+6 i }$
$\therefore \quad x+ i y=\frac{(2+3 i )(5-6 i )}{(5+6 i )(5-6 i )}$
$=\frac{10-12 i +15 i -18 i ^2}{25-36 i ^2}$
$=\frac{10+3 i -18(-1)}{25-36(-1)} \quad \ldots\left[\because i ^2=-1\right]$
$\therefore \quad x+ i y=\frac{28+3 i }{61}=\frac{28}{61}+\frac{3}{61} i $
Equating real and imaginary parts, we get
$x=\frac{28}{61} \text { and } y=\frac{3}{61}$
Solve the following quadratic equations. $i x^2-4 x-4 i=0$
Answer
$ i x^2-4 x-4 i=0 $ Multiplying throughout by $\mathrm{i}$, we get $ \begin{aligned} & \mathrm{i}^2 \mathrm{x}^2-4 \mathrm{ix}-4 \mathrm{i}^2=0 \\ & \therefore-\mathrm{x}^2-4 \mathrm{ix}+4=0 \ldots \ldots\left[\because \mathrm{i}^2=-1\right] \\ & \therefore \mathrm{x}^2+4 \mathrm{ix}-4=0 \end{aligned} $ Comparing with $a x^2+b x+c=0$, we get $ a=1, b=4 i, c=-4 $ Discriminant $=b^2-4 a c$ $ \begin{aligned} & =(4 i)^2-4 \times 1 \times-4 \\ & =16 i^2+16 \\ & =-16+16 \ldots . .\left[\because i^2=-1\right] \\ & =0 \end{aligned} $ So, the given equation has equal roots. These roots are given by $ \begin{aligned} x & =\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}} \\ & =\frac{-4 \mathrm{i} \pm \sqrt{0}}{2(1)}=\frac{-4 \mathrm{i}}{2} \end{aligned} $ $x=-2 \mathrm{i}$ $\therefore$ the roots of the given equation are $-2 \mathrm{i}$ and $-2 \mathrm{i}$.
Solve the following quadratic equations. $\mathrm{x}^2+4 \mathrm{ix}-4=0$
Answer
Given equation is $x^2+4 i x-4=0$ Comparing with $a x^2+b x+c=0$, we get $ \begin{aligned} & a=1, b=4 i, c=-4 \\ & \text { Discriminant }=b^2-4 a c \\ & =(4 i)^2-4 \times 1 \times-4 \\ & =16 i^2+16 \\ & \left.=-16+16 \ldots . . \because i^2=-1\right] \\ & =0 \end{aligned} $ So, the given equation has equal roots. These roots are given by $ \begin{aligned} x & =\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a} \text { }} \\ & =\frac{-4 \mathrm{i} \pm \sqrt{0}}{2(1)}=\frac{-4 \mathrm{i}}{2} \\ x & =-2 \mathrm{i} \end{aligned} $ $\therefore$ the roots of the given equation are $-2 \mathrm{i}$ and $-2 \mathrm{i}$.
Solve the following quadratic equations. $2 \mathrm{x}^2+3 \mathrm{ix}+2=0$
Answer
$2 \mathrm{x}^2+3 \mathrm{ix}+2=0$ Solution: Given equation is $2 x^2+3 i x+2=0$ Comparing with $a x^2+b x+c=0$, we get $ a=2, b=3 i, c=2 $ Discriminant $=b^2-4 a c$ $ \begin{aligned} & =(3 i)^2-4 \times 2 \times 2 \\ & =9 i^2-16 \\ & =-9-16 \\ & =-25<0 \end{aligned} $ So, the given equation has complex roots. These roots are given by $ \begin{aligned} x & =\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 a \mathrm{a}}}{2 \mathrm{a}} \\ & =\frac{-3 \mathrm{i} \pm \sqrt{-25}}{2(2)} \\ x & =\frac{-3 \mathrm{i} \pm 5 \mathrm{i}}{4} \\ x & =\frac{-3 \mathrm{i}+5 \mathrm{i}}{4} \text { or } x=\frac{-3 \mathrm{i}-5 \mathrm{i}}{4} \\ x & =\frac{1}{2} \mathrm{i} \text { } x=-2 \mathrm{i} \end{aligned} $ $\therefore$ the roots of the given equation are $\frac{1}{2} \mathrm{i}$ and $-2 \mathrm{i}$.
Solve the following quadratic equations. $x^2-4 x+13=0$
Answer
Given equation is $x^2-4 x+13=0$ Comparing with $a x^2+b x+c=0$, we get $ a=1, b=-4, c=13 $ Discriminant $=b^2-4 a c$ $ \begin{aligned} & =(-4)^2-4 \times 1 \times 13 \\ & =16-52 \\ & =-36<0 \end{aligned} $ So, the given equation has complex roots. These roots are given by $ \begin{aligned} x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ & =\frac{-(-4) \pm \sqrt{-36}}{2(1)} \\ & =\frac{4 \pm 6 \mathrm{i} \text { }}{2}=2 \pm 3 \mathrm{i} \end{aligned} $ $\therefore$ the roots of the given equation are $2+3 \mathrm{i}$ and $2-3 \mathrm{i}$.
Solve the following quadratic equations. $3 x^2-7 x+5=0$
Answer
Given equation is $3 x^2-7 x+5=0$ Comparing with $a x^2+b x+c=0$, we get $ a=3, b=-7, c=5 $ Discriminant $=b^2-4 a c$ $ \begin{aligned} & =(-7)^2-4 \times 3 \times 5 \\ & =49-60 \\ & =-11<0 \end{aligned} $ So, the given equation has complex roots. These roots are given by $ \begin{aligned} x & =\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}} \\ & =\frac{-(-7) \pm \sqrt{-11}}{2(3)} \\ x & =\frac{7 \pm \sqrt{11} \mathrm{i}}{6} \end{aligned} $
$\therefore$ the roots of the given equation are $\frac{7+\sqrt{11} i}{6}$ and $\frac{7-\sqrt{11} i}{6}$
Solve the following quadratic equations. $2 x^2-\sqrt{ } 3 x+1=0$
Answer
Given equation is $2 x^2-\sqrt{3} x+1=0$ Comparing with $a x^2+b x+c=0$, we get $ a=2, b=-\sqrt{3}, c=1 $ Discriminant $=b^2-4 a c$ $ \begin{aligned} & =(-\sqrt{ } 3)^2-4 \times 2 \times 1 \\ & =3-8 \\ & =-5<0 \end{aligned} $ So, the given equation has complex roots. These roots are given by $ \begin{aligned} x & =\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}} \\ & =\frac{-(-\sqrt{3}) \pm \sqrt{-5}}{2(2)} \\ x & =\frac{\sqrt{3} \pm \sqrt{5} \mathrm{i}}{4} \text { } \end{aligned} $ $\therefore$ the roots of the given equation are $\frac{\sqrt{3}+\sqrt{5} i}{4}$ and $\frac{\sqrt{3}-\sqrt{5} i}{4}$
Solve the following quadratic equations. $8 x^2+2 x+1=0$
Answer
Given equation is $8 x^2+2 x+1=0$ Comparing with $a x^2+b x+c=0$, we get $ \begin{aligned} & a=8, b=2, c=1 \\ & \text { Discriminant }=b^2-4 a c \\ & =(2)^2-4 \times 8 \times 1 \\ & =4-32 \\ & =-28<0 \end{aligned} $ So, the given equation has complex roots. These roots are given by $ \begin{aligned} x & =\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}} \\ & =\frac{-2 \pm \sqrt{-28}}{2(8)} \\ & =\frac{-2 \pm 2 \sqrt{7} \mathrm{i}}{16 \text { }} \\ x & =\frac{-1 \pm \sqrt{7} \mathrm{i}}{8} \end{aligned} $ $\therefore$ the roots of the given equation are $\frac{-1+\sqrt{7 \mathrm{i}}}{8}$ and $\frac{-1-\sqrt{7 \mathrm{i}}}{8}$