Find X
43x^3+13x^2+18x = 456
2x(3x+2)^2 = 312
To find the value of x in these equations, we'll start by rearranging the equations to isolate x.
1. Equation 1: 43x^3 + 13x^2 + 18x = 456
We'll set the equation equal to zero by subtracting 456 from both sides:
43x^3 + 13x^2 + 18x - 456 = 0
Now, we'll factor out any common factors from the expression:
x^2(43x + 13) + 18(x - 8) = 0
To solve this equation, we'll use factoring. Since the product of two factors is zero, at least one of the individual factors must be zero.
Setting the first factor to zero:
x^2 = 0
x = 0
Setting the second factor to zero:
43x + 13 = 0
43x = -13
x = -13/43
So, the solutions for Equation 1 are x = 0 and x = -13/43.
2. Equation 2: 2x(3x + 2)^2 = 312
Divide both sides of the equation by 2 to simplify:
x(3x + 2)^2 = 156
Taking the square root of both sides gives:
√[x(3x + 2)^2] = √156
(x)(3x + 2) = √156
Expanding the equation gives:
3x^2 + 2x = ±√156
Rearrange it to set it equal to zero by subtracting √156 from both sides:
3x^2 + 2x - √156 = 0
This equation does not easily factor, so we can use the quadratic formula to find the value of x:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 3, b = 2, and c = -√156.
Plugging these values into the quadratic formula,
x = (-2 ± √(2^2 - 4(3)(-√156))) / (2(3))
x = (-2 ± √(4 + 12√39)) / 6
x = (-2 ± √(4 + 12 * 3.13)) / 6
x = (-2 ± √(4 + 37.56)) / 6
x = (-2 ± √41.56) / 6
So, the solutions for Equation 2 are x ≈ (-2 + √41.56)/6 and x ≈ (-2 - √41.56)/6.
Therefore, the solutions for both equations are:
Equation 1: x = 0 and x = -13/43
Equation 2: x ≈ (-2 + √41.56)/6 and x ≈ (-2 - √41.56)/6.