Question

Show That the roots of the equation x²-4x+13=0 are conjugate complex numbers solve this question

Answer 1

$\huge{\sf{Answer:}}$

Given Equation,

$\mathtt{x {}^{2} - 4x + 13 = 0}$

On comparing with ax²+bx+c = 0,

$\mathtt{a = 1 \: b = - 4 \: and \: c \: = 13}$

Now,

Let us find the discriminant of the above equation

•Discriminant of a quadratic equation,

$\sf{d = b {}^{2} - 4ac} \\ \\ \rightarrow \: \sf{d = ( - 4) {}^{2} - 4(1)(13) } \\ \\ \rightarrow \sf{d = 16 - 52} \\ \\ \rightarrow \: \boxed{\sf{d = - 36}}$

•Roots of x would be:

$\sf{x = \frac{ - b \pm \sqrt{d} }{2a} } \\ \\ \implies \: \sf{x = \frac{ - ( - 4) \pm \sqrt{ - 36} }{2}} \\ \\ \implies \sf{x = \frac{4 \pm \sqrt{36}i }{2} } \\ \\ \implies \: \sf{x = \frac{4 \pm \: 6i}{2} } \\ \\ \implies \: \boxed{\sf{x = 2 \pm \: 3i}}$

• Thus,one root of the equation is conjugate to the other