1.Solve: x(x2 + 16) = 0
2.Solve: 2x3 + 6x2 – x – 3 = 0
3. Solve: 2x = x2
1. x = 0, 4i or -4i
(i is the square root of -1)
2. One of then roots is -3. I got that by trying small integers. That means (x+3) is a factor. Divide x+3 into
2x^3 + 6x2 – x – 3 and you get the other factor, 2x^2-1. That can also be factored, giving you the other two solutions.
3. Rewrite it as x(x-2) = 0
That's an easy one. What are the two x values that satisfy that equation?
I assume x2 means x^2 ---- to the second power
1.Solve: x(x2 + 16) = 0
If x is zero, the equation is true, so x = 0 is a solution.
The other solutions are from x^2+16 = 0
x^2 = -16
x = + or - sqrt (-16)
sqrt (-16) = 4 sqrt (-1)
sqrt (-1) is usually called i
so x = + or - 4i
2.Solve: 2x3 + 6x2 – x – 3 = 0
factor out (x+3)
( I guessed x + 3 or x-3 would be factors because of that lonely 3 at the end)
(x+3)(2x^2-1)) = 0
x = -3 is a solution
two others come from 2x^2-1 = 0
x^2 = 1/2
x = + or - (1/2) sqrt( 2)
3. Solve: 2x = x2
that is x^2 - 2x = 0
which is
x (x-2) = 0
x = 0 or x = 2
To solve these equations, we will use algebraic methods.
1. Solve: x(x^2 + 16) = 0
Firstly, we can apply the zero product property, which states that if a product of factors is equal to zero, then at least one of the factors must be zero. Therefore, our equation can be written as two separate equations:
x = 0 (Equation 1)
or
x^2 + 16 = 0 (Equation 2)
Equation 1 is straightforward, where x = 0.
To solve Equation 2, we subtract 16 from both sides:
x^2 = -16
Taking the square root of both sides, we get:
x = ±√(-16)
Since the square root of a negative number is not a real number, this equation has no real solutions.
Therefore, the solutions to the original equation x(x^2 + 16) = 0 are x = 0 only.
2. Solve: 2x^3 + 6x^2 - x - 3 = 0
In this case, it is not immediately obvious how to factor the equation. We can try factoring by grouping or synthetic division, but it would involve some trial and error. However, we can also use a numerical method called the Newton-Raphson method or graphical methods to find approximate solutions.
The Newton-Raphson method involves iterative calculations and can provide a more accurate solution. However, it is a bit complex to explain fully here. Graphical methods involve plotting the equation on a graph and finding the x-intercepts (where the graph intersects the x-axis).
3. Solve: 2x = x^2
To solve this equation, we need to rearrange it into a quadratic equation form:
x^2 - 2x = 0
Next, we factor x out:
x(x - 2) = 0
Applying the zero product property, we have two separate equations:
x = 0 (Equation 1)
or
x - 2 = 0 (Equation 2)
Solving Equation 2, we add 2 to both sides:
x = 2
Therefore, the solutions to the equation 2x = x^2 are x = 0 and x = 2.