1. solve: x^3 =4x
2. solve: (x-3)^2 = -4
3. solve: (6 +5i) ^2
4.Two integers have a sum of -4 The sum of their squares is 40. What are the two integers?
1. x(x^2-4) = 0
x(x+2)(x-2) = 0
Do you see the three possibilties now?
2. Take the square root of both sides and then do some algebra to get x
3. There is nothing to solve. You do not have an equation.
4. Call the numbers x and y
x + y = -4, so y = -4-x
Then solve the equation
x^2 + (-4-x)^2 = 40
2x^2 +8x -24 = 0
x^2 +4x -12 = 0
(x+6)(x-2) = 0
I will do number 2 only.
We have this:
(x-3)^2 = -4
The left side becomes x^2 - 6x + 9.
We now have this:
x^2 - 6x + 9 = -4
Add 4 to both sides and then equate to zero.
x^2 - 6x + 9 + 4 = 0
x^2 - 6x + 13 = 0
This quadratic equation cannot be solved by factoring. In that case, use the quadratic formula to find x.
If you do not know how to use the quadratic formula, write back.
1. To solve the equation x^3 = 4x, we need to find the values of x that satisfy the equation. We'll start by moving all terms to one side of the equation:
x^3 - 4x = 0
Now, we factor out x:
x(x^2 - 4) = 0
From here, we have two separate equations:
x = 0
x^2 - 4 = 0
For the first equation, x = 0 is one solution.
For the second equation, we can solve it by factoring:
(x + 2)(x - 2) = 0
This gives us two more solutions: x = -2 and x = 2.
Therefore, the solutions to the equation x^3 = 4x are x = 0, x = -2, and x = 2.
2. To solve the equation (x-3)^2 = -4, we'll start by moving the constant term to the other side:
(x-3)^2 + 4 = 0
Next, we'll take the square root of both sides:
sqrt((x-3)^2 + 4) = sqrt(0)
(x-3) ± sqrt(4) = ±2
Now, we can simplify:
(x-3) ± 2 = ±2
To solve for x, we need to consider both cases of adding and subtracting 2:
Case 1: (x-3) + 2 = 2
Simplifying the equation gives us x - 1 = 2, so x = 3.
Case 2: (x-3) - 2 = -2
Simplifying the equation gives us x - 5 = -2, so x = 3.
Therefore, the solution to the equation (x-3)^2 = -4 is x = 3.
3. To solve the equation (6 + 5i)^2, we'll start by expanding the square:
(6 + 5i)^2 = (6 + 5i)(6 + 5i)
Using the FOIL method, we get:
(6 * 6) + (6 * 5i) + (5i * 6) + (5i * 5i)
Simplifying further gives us:
36 + 30i + 30i + 25i^2
Since i^2 equals -1, we can substitute it:
36 + 30i + 30i + 25(-1) = 36 + 60i - 25 = 11 + 60i
Therefore, the solution to the equation (6 + 5i)^2 is 11 + 60i.
4. Let's solve the problem using algebraic equations.
Let's consider the two integers as x and y.
According to the given information, the sum of the two integers is -4, so we can write the equation:
x + y = -4
The sum of their squares is given as 40, so we can write a second equation:
x^2 + y^2 = 40
To solve these two equations simultaneously, we can use the method of substitution or elimination.
Let's use the method of substitution. Starting with the first equation, we can solve it for one variable:
x = -4 - y
Now, substitute this expression for x in the second equation:
(-4 - y)^2 + y^2 = 40
Expanding and simplifying the equation gives us:
16 + 8y + y^2 + y^2 = 40
2y^2 + 8y + 16 = 40
Rearranging and simplifying further:
2y^2 + 8y - 24 = 0
Divide the equation by 2:
y^2 + 4y - 12 = 0
Now, factorize the equation:
(y + 6)(y - 2) = 0
This gives us two possible values for y:
y1 = -6
y2 = 2
Substituting each value into the expression for x, we get:
x1 = -4 - (-6) = -4 + 6 = 2
x2 = -4 - 2 = -6
Therefore, the two integers that satisfy the given conditions are (2, -6) and (-6, 2).