The equation $x^2-9=0$ has 2 real solutions.
The equation $x^3+3 x^2+5 x+15=0$ has $\square$ real and $\square$ imaginary solution(s).
Answer (1)
Factor the cubic equation: x 3 + 3 x 2 + 5 x + 15 = ( x + 3 ) ( x 2 + 5 ) .
Solve for x when x + 3 = 0 , which gives the real solution x = − 3 .
Solve for x when x 2 + 5 = 0 , which gives the imaginary solutions x = ± i 5 .
The equation has 1 real and 2 imaginary solutions: 1 real and 2 imaginary.
Explanation
Problem Analysis We are given the equation x 3 + 3 x 2 + 5 x + 15 = 0 and we need to find the number of real and imaginary solutions.
Factoring the Equation First, let's factor the given cubic equation. We can rewrite the equation as x 2 ( x + 3 ) + 5 ( x + 3 ) = 0 .
Further Factoring Now, we can factor out ( x + 3 ) from the equation, which gives us ( x + 3 ) ( x 2 + 5 ) = 0 .
Finding the Solutions To find the solutions, we set each factor equal to zero. So, we have x + 3 = 0 and x 2 + 5 = 0 .
Real Solution From x + 3 = 0 , we get x = − 3 . This is a real solution.
Imaginary Solutions From x 2 + 5 = 0 , we get x 2 = − 5 . Taking the square root of both sides, we have x = ± − 5 = ± i 5 . These are two imaginary solutions.
Conclusion Therefore, the equation x 3 + 3 x 2 + 5 x + 15 = 0 has 1 real solution and 2 imaginary solutions.
Examples
Understanding the nature of polynomial roots (real or imaginary) is crucial in many engineering applications, such as control systems design. For instance, when designing a feedback control system, the roots of the characteristic equation determine the stability of the system. Real roots correspond to stable or unstable behaviors, while complex roots indicate oscillatory behaviors. By analyzing the roots, engineers can tune the system parameters to ensure stability and desired performance. This ensures that systems like cruise control in cars or robotic arms in manufacturing plants operate smoothly and predictably.
