Question

Find the complex roots and possible number of real and imaginary roots for the equations. Then find all roots . Show Steps to Solve

X^2(x+2)(x-5)=0
And
(X^2+9)(x^2+6)=0

Answer 1

For the equation X^2(x+2)(x-5)=0, there are three real roots: 0, -2, and 5. For the equation (X^2+9)(x^2+6)=0, all four roots are imaginary: ± 3i and ± √6 i.

Step-by-step explanation:

To find the complex roots and the possible number of real and imaginary roots for the given equations, we will solve each factor set to zero separately.

Finding roots for X2(x+2)(x-5)=0

This equation is factored completely, so we set each factor equal to zero and solve:

• X2 = 0 → X = 0 (this is a double root)
• x+2 = 0 → X = -2
• x-5 = 0 → X = 5

Here we have three real roots: 0 (repeated twice), -2, and 5. There are no imaginary roots in this equation.

Finding roots for (X2+9)(x2+6)=0

For the equation (X2+9)(x2+6)=0, we solve each quadratic factor separately:

• X2+9 = 0 → X2 = -9 → X = ± 3i (imaginary roots)
• x2+6 = 0 → X2 = -6 → X = ± √6 i (imaginary roots)

For this equation, there are no real roots; all roots are imaginary: 3i, -3i, √6i, and -√6i.