Question

2.Solve the equation (i)x^(3)+6x+20=0 ​

Answer 1

Answer:

Since all coefficients of [math]x^3 + 6x + 20 = 0[/math] are real numbers, if one root is [math]1+3i[/math], then second root will be [math]1 - 3i[/math].

Cubic polynomial will have at least one real root and let that be ‘[math]a[/math]’.

We can find the 3rd root in one of the following ways:

Sum of the roots =[math] a + 1 + 3i + 1 - 3i = -[/math] Coefficient of [math]x^2 = 0[/math]

[math]=> a + 2 = 0 => a = -2[/math]

Product of the roots =[math] (a)(1+3i)(1–3i) = -[/math] Coefficient of x^0 or [math](-1)* [/math]Constant

[math]=> a (1 - 9i^2) = -20 => a (10) = - 20[/math]

[math]=> a = -2[/math]

Roots are:[math] (1+3i), (1–3i), (-2)[/math]

Answer 2

Answer:

3x+6x+20=0

9x+20=0

9x=0-20

9x=20

x=20/9

x= 2.2

Step-by-step explanation:

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