The solution of [tex]x^{2}-2 x+5=0[/tex] are 1 + 2i and 1 – 2i
Solution:
Given, equation is [tex]x^{2}-2 x+5=0[/tex]
We have to find the roots of the given quadratic equation
Now, let us use the quadratic formula
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex] --- (1)
Let us determine the nature of roots:
Here in [tex]x^{2}-2 x+5=0[/tex] a = 1 ; b = -2 ; c = 5
[tex]b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16[/tex]
Since [tex]b^2 - 4ac < 0[/tex] , the roots obtained will be complex conjugates.
Now plug in values in eqn 1, we get,
[tex]x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}[/tex]
On solving we get,
[tex]x=\frac{2 \pm \sqrt{4-20}}{2}[/tex]
[tex]x=\frac{2 \pm \sqrt{-16}}{2}[/tex]
[tex]x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}[/tex]
we know that square root of -1 is "i" which is a complex number
[tex]\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}[/tex]
Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i
Answer:
1 + 2i and 1 – 2i
Step-by-step explanation:
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