How would you solve the following I'm stumped..
1. 2x^2(squared)=30-4x
2. X^3(cubed)+8x=-6x^2(squared)
Thanks!
First, set each equation equal to 0:
1. 2x^2 + 4x - 30 = 0
2. x^3 + 6x^2 + 8x = 0
Try to factor when possible:
1. (2x - 6)(x + 5) = 0
2. x(x^2 + 6x + 8) = 0
x(x + 2)(x + 4) = 0
Set each factor equal to 0:
1. 2x - 6 = 0
x + 5 = 0
2. x = 0
x + 2 = 0
x + 4 = 0
Can you take it from here to find the solutions for x in both problems?
I hope this explanation helps.
Yes thank you!
To solve the given equations, we need to find the values of x that satisfy each equation. Let's go through the steps for each equation:
1. 2x^2 = 30 - 4x:
Step 1: Move all the terms to one side to set the equation to zero:
2x^2 + 4x - 30 = 0
Step 2: Simplify the equation if possible:
Divide the entire equation by 2 to simplify it:
x^2 + 2x - 15 = 0
Step 3: Factoring or using the quadratic formula to solve for x:
In this case, we can factor the quadratic equation:
(x + 5)(x - 3) = 0
Step 4: Set each factor equal to zero and solve for x:
x + 5 = 0 ---> x = -5
x - 3 = 0 ---> x = 3
So, the solutions for the equation 2x^2 = 30 - 4x are x = -5 and x = 3.
2. x^3 + 8x = -6x^2:
Step 1: Move all the terms to one side to set the equation to zero:
x^3 + 6x^2 + 8x = 0
Step 2: Simplify the equation if possible:
Factor out an x:
x(x^2 + 6x + 8) = 0
Step 3: Factor or use other methods to solve for x:
(x)(x + 4)(x + 2) = 0
Step 4: Set each factor equal to zero and solve for x:
x = 0, x + 4 = 0 ---> x = -4
x + 2 = 0 ---> x = -2
So, the solutions for the equation x^3 + 8x = -6x^2 are x = 0, x = -4, and x = -2.
Remember to always check your solutions by substituting them back into the original equations to ensure they satisfy the given equations.