Question #b9b3d

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Answer 1 · 2016-10-30T20:06:39

$p = - \frac{1}{2} \pm i$ $p=-1/2+-i$

Standard form: $y = a x^{2} + b x + c \Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $" "y=ax^2+bx+c" " =>" " x=(-b+-sqrt(b^2-4ac))/(2a)$

In this $a = 1; b = 1; c = 4$ $" "a=1"; "b=1"; "c=4$ giving

$p = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (4)}}{2 (1)}$ $p=(-1+-sqrt(1^2-4(1)(4)))/(2(1))$

$p = - \frac{1}{2} \pm \frac{\sqrt{- 16}}{2}$ $p=-1/2+-sqrt(-16)/2$

Which is the same as:

$p = - \frac{1}{2} \pm \frac{\sqrt{4^{2} \times (- 1)}}{2}$ $p=-1/2+-sqrt(4^2xx(-1))/2$

$p=-1/2+-(cancel(2)sqrt(-1))/(cancel(2))$

But $sqrt(-1) = i$

$p=-1/2+-i$