How do you solve the quadratic with complex numbers given $x^{2} - 4 x + 5 = 0$ $x^2-4x+5=0$ ?

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How do you solve the quadratic with complex numbers given $x^{2} - 4 x + 5 = 0$ $x^2-4x+5=0$ ?

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Answer 1 · 2016-08-28T15:11:21

$x = 2 + i$ $x=2+i$ and $x = 2 - i)$ $x=2-i)$

Explanation:

For the equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , the roots are given by $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

It is apparent that if the discriminant $b^{2} - 4 a c < 0$ $b^2-4ac<0$ , we have complex roots.

In the equation $x^{2} - 4 x + 5 = 0$ $x^2-4x+5=0$ , the discriminant is ${(- 4)}^{2} - 4 \times 1 \times 5 = 16 - 20 = - 4 < 0$ $(-4)^2-4×1×5=16-20=-4<0$ and hence roots are complex.

These are $x = \frac{- (- 4) \pm \sqrt{- 4}}{2 \times 1}$ $x=(-(-4)+-sqrt(-4))/(2×1)$

= $\frac{4 \pm 2 i}{2}$ $(4+-2i)/2$ i.e.

$x = 2 + i$ $x=2+i$ and $x = 2 - i$ $x=2-i$

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How do you solve the quadratic with complex numbers given $x^{2} - 4 x + 5 = 0$ $x^2-4x+5=0$ ?

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