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Find the solution set of the quadratic equation over the set of complex numbers. 5x2 + 8x + 4 = 0 A) x = −1 − i 2 or −1 + i 2 B) x = − 4 5 − 2i 5 or − 4 5 + 2i 5 C) x = − 2 5 − 6i 5 or − 2 5 + 6i 5 D) x = − 1 4 − 3i 4 or − 1 4 + 3i 4

Respuesta :

altavistard
5x^2 + 8x + 4 = 0  is a quadratic.  a=5, b=8 and c=4

Thus,

        -8 plus or minus sqrt( 8^2-4(5)(4) )
 x = ------------------------------------------------
                           2(5)
         -8 plus or minus sqrt(64-80)
    = ---------------------------------------------
                                10


         -8 plus or minus sqrt(-16)          -8 plus or minus i*4
    = -------------------------------------- = -----------------------------
                                10                                     10
           -4 plus or minus i*2
     = ------------------------------    Can you now choose the correct answer?
                         5

Answer:

The solutions to the quadratic equation are:

[tex]x=-\frac{4}{5}+i\frac{2}{5},\:x=-\frac{4}{5}-i\frac{2}{5}[/tex]

Step-by-step explanation:

Complex numbers are numbers of the form [tex]a+bi[/tex], where [tex]a[/tex] and [tex]b[/tex] are real numbers.

For a quadratic equation of the form [tex]ax^2+bx+c=0[/tex] the solutions are

[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

If [tex]b^2-4ac<0[/tex], the equation has two complex solutions that are not real.

Quadratic equations with a negative discriminant have no real number solution. However, if we extend our number system to allow complex numbers, quadratic equations will always have a solution.

This quadratic equation [tex]5x^2\:+\:8x\:+\:4\:=\:0[/tex] is not factorable, so we apply the quadratic formula.

[tex]\mathrm{For\:}\quad a=5,\:b=8,\:c=4\\\\x_{1,\:2}=\frac{-8\pm \sqrt{8^2-4\cdot \:5\cdot \:4}}{2\cdot \:5}\\[/tex]

[tex]x_1=\frac{-8+\sqrt{8^2-4\cdot \:5\cdot \:4}}{2\cdot \:5}=\frac{-8+\sqrt{16}i}{2\cdot \:5}=-\frac{4}{5}+\frac{2}{5}i[/tex]

[tex]x_2=\frac{-8-\sqrt{8^2-4\cdot \:5\cdot \:4}}{2\cdot \:5}=\frac{-8-\sqrt{16}i}{10}=-\frac{4}{5}-\frac{2}{5}i[/tex]

The solutions to the quadratic equation are

[tex]x=-\frac{4}{5}+i\frac{2}{5},\:x=-\frac{4}{5}-i\frac{2}{5}[/tex]

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