Question

1)3x^3-12x=0; 3)x^3-5x^2-x+5=0

Answer 1

1) 3x(x²-4)=0
3x=0
x=0

x²=4
x=2 x=-2

2) x(x²-1) -5(x²-1)=0
(x²-1)(x-5)=0
x²-1=0
x=1 x=-1

x-5=0
x=5

Answer 2

To solve the given equations, we can use algebraic methods to find the values of 'x' that satisfy each equation.

1. $3x^3 - 12x = 0$

Step 1: Factor out 'x' from the equation. $x(3x^2 - 12) = 0$

Step 2: Now set each factor to zero and solve for 'x'.

From the first factor: $x = 0$

From the second factor: $3x^2 - 12 = 0$

Step 3: Divide the equation by 3 to simplify. $x^2 - 4 = 0$

Step 4: Factor the quadratic equation. $(x - 2)(x + 2) = 0$

Step 5: Set each factor to zero and solve for 'x'. $x - 2 = 0$ or $x + 2 = 0$

Solutions: $x = 0$, $x = 2$, or $x = -2$

2. $x^3 - 5x^2 - x + 5 = 0$

This equation is a cubic equation, and it's not easy to solve by factoring. We can use numerical methods or approximate solutions.

One method to find approximate solutions is the Newton-Raphson method. Let's use the initial guess $x_0 = 1$ to find one root:

Step 1: Find the derivative of the equation with respect to 'x'. $f(x) = x^3 - 5x^2 - x + 5$ $f'(x) = 3x^2 - 10x - 1$

Step 2: Apply the Newton-Raphson formula to find the next approximation $x_1$: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$ $x_1 = 1 - \frac{1^3 - 5(1)^2 - 1 + 5}{3(1)^2 - 10(1) - 1}$ $x_1 = 1 - \frac{-1}{-8} = 1 + \frac{1}{8} = \frac{9}{8} = 1.125$

So, one root of the equation is approximately $x \approx 1.125$.

To find the other roots, you can repeat the Newton-Raphson method using different initial guesses or use other numerical methods like the bisection method or the secant method.

Please note that for the second equation, there may be one or two more real roots. In some cases, there might also be complex roots. Numerical methods will help us find approximate solutions for such cases.