How do you solve the quadratic with complex numbers given $5 x^{2} + 8 x + 5 = 0$ $5x^2+8x+5=0$ ?

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Answer 1 · 2016-09-04T09:48:01

$x = - \frac{4}{5} - i \frac{3}{5}$ $x=-4/5-i3/5$ and $x = - \frac{4}{5} + i \frac{3}{5}$ $x=-4/5+i3/5$

Quadratic formula gives the solution of quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ as

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

Hence solution of $5 x^{2} + 8 x + 5 = 0$ $5x^2+8x+5=0$ is given by

$x = \frac{- 8 \pm \sqrt{8^{2} - 4 \times 5 \times 5}}{2 \times 5}$ $x=(-8+-sqrt(8^2-4×5×5))/(2×5)$

= $\frac{- 8 \pm \sqrt{64 - 100}}{10}$ $(-8+-sqrt(64-100))/10$

= $\frac{- 8 \pm \sqrt{- 36}}{10}$ $(-8+-sqrt(-36))/10$

= $\frac{- 8 \pm i 6}{10} = - \frac{4}{5} \pm i \frac{3}{5}$ $(-8+-i6)/10=-4/5+-i3/5$

i.e. $x = - \frac{4}{5} - i \frac{3}{5}$ $x=-4/5-i3/5$ and $x = - \frac{4}{5} + i \frac{3}{5}$ $x=-4/5+i3/5$

How do you solve the quadratic with complex numbers given 5x^2+8x+5=05x2+8x+5=05x^2+8x+5=0?