Question

Solve the following quadratic equations by factorization method:
(i) $x^2-5ix-6=0$
(ii) $x^2+4ix-4=0$

Answer 1

$\left(i\right)$ $x^2-5ix-6=0$

$\Rightarrow$ Here, $a=1,b=-5i,c=-6$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-(-5i)\pm \sqrt{(-5i{)}^2-4(1\left)\right(-6)}}{2\left(1\right)}$

$=\frac{5i\pm \sqrt{25i^2+24}}{2}$

$=\frac{5i\pm \sqrt{-25+24}}{2}$ [ Since, $i^2=-1$ ]

$=\frac{5i\pm \sqrt{-1}}{2}$

$=\frac{5i\pm i}{2}$ [ Since, $i=\sqrt{-1}$ ]

$x=\frac{5i+i}{2}$ and $x=\frac{5i-i}{2}$

$\therefore$ $x=3i$ and $x=2i$

$\left(ii\right)$ $x^2+4ix-4=0$

$\Rightarrow$ Here, $a=1,b=4i,c=-4$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-4i\pm \sqrt{(4i{)}^2-4(1\left)\right(-4)}}{2\left(1\right)}$

$=\frac{-4i\pm \sqrt{16i^2+16}}{2}$

$=\frac{-4i\pm \sqrt{-16+16}}{2}$

$=\frac{-4i}{2}$

$\therefore$ $x=\frac{-4i}{2}=-2i$