Solve the following quadratic equations. x^2-4 x+13=0 - Vidyadip

Question

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 $2-3 \mathrm{i}$.

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