Show the following quadratic equation: 2x^2+15 ix-i=0 - Vidyadip

Question

Show the following quadratic equation:
$2\text{x}^2+\sqrt{15}\ \text{ix}-\text{i}=0$

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Answer

$2 x^2+\sqrt{15} ix-i=0$
Comparing the given Equation with the general form $a x^2+b x+c=0$,
we get $a=2, b=\sqrt{15} i, c=-i$
Substituting a and in.
$\alpha=\frac{-b+\sqrt{b^2-4 ac}}{2 a} \text { and } \beta=\frac{-b-\sqrt{b^2-4 ac}}{2 a}$
$a=\frac{-\sqrt{15} i+\sqrt{-15+8 i}}{4} \text { and } \beta=\frac{-\sqrt{15} i-\sqrt{-15+8} i}{4}$
Let $\sqrt{-15+8 i }= a + bi$
$\Rightarrow-15+8 i=(a+bi)^2$
$\Rightarrow-15+8 i=a^2-b^2+2 abi$
$\Rightarrow a^2-b^2=-15 and 2 abi=8 i$
$\text { Now }\left(a^2+b^2\right)=\left(a^2-b^2\right)+4 a^2 b^2$
$\Rightarrow\left(a^2+b^2\right)=(15)^2+64=289$
$\Rightarrow a^2+b^2=17$
Solving $a^2-b^2=-15$ and $a^2+b^2=17$, we get
$a^2=1 \text { and } b^2=16$
$\Rightarrow a= \pm 1 \text { and } b= \pm 4$
$\Rightarrow a=1, b=4 \text { or } a=-1, b=-4$
$\therefore \sqrt{-15+8 i}=1+4 i,-1-4 i$
when $\sqrt{-15+8} i =1+4 i$
$\alpha=\frac{-\sqrt{15 i+1+4 i}}{4}=\frac{1+(4-\sqrt{15}) i}{4}$
and $\beta=\frac{-\sqrt{15 i}-(1+4 i )}{4}=\frac{-1-(4+\sqrt{15}) i }{4}$
When $\sqrt{-15+8 i }=-1-4 i$
$\alpha=\frac{-\sqrt{15 i}-1-4 i}{4}=\frac{-1-(4+\sqrt{15})}{4}$
and $\beta=\frac{\sqrt[4]{15 i}-(-1-4 i )}{4}=\frac{4+(4-\sqrt{5}) i }{4}$

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