Solve the following quadratic equations. x^2-(2+i) x-(1-7 i)=0

Question

Solve the following quadratic equations.
$x^2-(2+i) x-(1-7 i)=0$

✓

Answer

Given equation is $x^2-(2+i) x-(1-7 i)=0$
Comparing with $a x^2+b x+c=0$, we get
$
a=1, b=-(2+i), c=-(1-7 i)
$
Discriminant $=b^2-4 a c$
$
\begin{aligned}
& =[-(2+i)]^2-4 \times 1 \times-(1-7 i) \\
& =4+4 i+i^2+4-28 i \\
& =4+4 i-1+4-28 i \ldots . . .\left[\because i^2=-1\right] \\
& =7-24 i
\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{-[-(2+i)] \pm \sqrt{7-24 i}}{2(1)} \\
& =\frac{(2+i) \pm \sqrt{7-24 i}}{2}
\end{aligned}
$

Let $\sqrt{7-24 i}=a+b i$, where $a, b \in R$
Squaring on both sides, we get
$
\begin{array}{r}
7-24 i=a^2+i^2 b^2+2 a b i \\
\therefore \quad 7-24 i=\left(a^2-b^2\right)+2 a b i
\end{array}
$
$
\ldots\left[\because i^2=-1\right]
$
Equating real and imaginary parts, we get
$
a^2-b^2-7 \text { and } 2 a b=-24
$
$
\begin{array}{ll}
\therefore & a^2-b^2-7 \text { and } b=\frac{-12}{a} \\
\therefore & a^2-\left(\frac{-12}{a}\right)^4=7 \\
\text { } \\
\therefore & a^2-\frac{144}{a^2}=7 \\
\therefore \quad & a^4-144=7 a^2 \\
\therefore \quad & a^4-7 a^2-144-0 \\
\therefore \quad & \left(a^2-16\right)\left(a^2-9\right)=0 \\
\therefore & a^2=16 \text { or } a^2-9
\end{array}
$
But $a \in R$
$
\begin{array}{ll}
\therefore & a^2 \neq-9 \\
\therefore & a^2=16 \\
\therefore & a= \pm 4
\end{array}
$
When $a-4, b-\frac{-12}{4}=-3$
When $a=-4, b=\frac{-12}{-4}=3$
$
\begin{array}{ll}
\therefore & \sqrt{7-24 i}- \pm(4-3 i) \\
\therefore & x=\frac{(2+i) \pm(4-3)}{2} \text { } \\
\therefore & x-\frac{(2+i)+(4-3)}{2} \text { or } x=\frac{(2+\mathrm{i})-(4-3 \mathrm{i})}{2} \\
\therefore & x=3-\mathrm{i} \text { of } x-1+2 \mathrm{i}
\end{array}
$