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Complex Numbers(p-1) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Complex Numbers(p-1)5 Marks
Question
Solve the following quadratic equations.
$x^2+3 i x+10=0$
✓

Answer

Given equation is $x^2+3 i x+10=0$
Comparing with $a \mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0$, we get
$
\begin{aligned}
& a=1, b=3 i, c=10 \\
& \text { Discriminant }=b^2-4 a c \\
& =(3 i)^2-4 \times 1 \times 10 \\
& =9 i^2-40 \\
& =-9-40 \ldots \ldots\left[\because i^2=-1\right] \\
& =-49
\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{-3 i \pm \sqrt{-49}}{2(1)} \\
& x=\frac{-3 i \pm 7 i}{2} \text { } \\
& x=\frac{-3 i+7 i}{2} \text { or } x=\frac{-3 i-7 i}{2} \\
& \therefore x=2 i \text { or } x=-5 i
\end{aligned}
$
$\therefore$ the roots of the given equation are $2 \mathrm{i}$ and $-5 \mathrm{i}$.
Check:
$
\begin{aligned}
& \text { If } x=2 i \text { and } x=-5 i \text { satisfy the given equation, then our answer is correct. } \\
& \text { L.H.S. }=x^2+3 i x+10 \\
& =(2 i)^2+3 i(2 i)+10 i \\
& =4 i^2+6 i^2+10 \\
& =10 i^2+10 \\
& =-10+10 \ldots . \cdot\left[i^2=-1\right] \\
& =0 \\
& =\text { R.H.S. } \\
& \text { L.H.S. }=x^2+3 i x+10 \\
& =(-5 i)^2+3 i(-5 i)+10 \\
& =25 i^2-15 i^2+10 \\
& =10 i^2+10 \\
& =-10+10 \ldots\left[-\ldots i^2=-1\right] \\
& =0 \\
& =\text { R.H.S. }
\end{aligned}
$
Thus, our answer is correct.

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