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How do you solve the quadratic with complex numbers given #-5x^2+12x-8=0#?

Answer 1

Lily Anderson

#x = 1.2 - 0.1 sqrt26 color(red)i, 1.2 + 0.1 sqrt26 color(red)i#

#-5x^2 + 12x - 8 = 0#

#x = (-b +- sqrt(b^2 - (4 a c))) / (2 * a)#

#a = -5, b = 12, c = -8#

#x = (-12 +- sqrt(12^2 - (4 * -5 * - 8))) / (2 * -5)#

#x = (-12 +- sqrt(144 - 160)) / (-10)#

#x = 1.2+- 0.1 sqrt(-26)#

#x = 1.2 +- 0.1 sqrt26 color(red)i#

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Answer 2

Madelyn Anderson

To solve the quadratic equation (-5x^2 + 12x - 8 = 0) using the quadratic formula, (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = -5), (b = 12), and (c = -8):

[x = \frac{{-12 \pm \sqrt{{12^2 - 4(-5)(-8)}}}}{{2(-5)}}]

[x = \frac{{-12 \pm \sqrt{{144 - 160}}}}{{-10}}]

[x = \frac{{-12 \pm \sqrt{{-16}}}}{{-10}}]

[x = \frac{{-12 \pm 4i}}{{-10}}]

[x = \frac{{-6 \pm 2i}}{{-5}}]

So the solutions are (x = \frac{{-6 + 2i}}{{-5}}) and (x = \frac{{-6 - 2i}}{{-5}}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.