how to find the x-interceps? x^4-29x^2+100?
x^4-29x^2+100
=(x^2 - 25)(x^2 - 4)
=(x-5)(x+5)(x-2)(x+2)
Does that help?
thanks
To find the x-intercepts of the equation x^4 - 29x^2 + 100, follow these steps:
Step 1: Set the equation equal to zero: x^4 - 29x^2 + 100 = 0.
Step 2: Factor the equation, if possible. Unfortunately, the given equation cannot be easily factored, so we need to use an alternative method.
Step 3: Make a substitution. Let's do a substitution: let u = x^2. Now the equation becomes u^2 - 29u + 100 = 0.
Step 4: Factor the substituted equation. Factor the equation u^2 - 29u + 100 = 0 to determine the possible values of u. In this case, the factors cannot be easily found, so we will proceed to the next step.
Step 5: Use the quadratic formula. Apply the quadratic formula to solve for u in the equation u^2 - 29u + 100 = 0. The quadratic formula is given by u = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = -29, and c = 100.
To find the values of u, substitute the corresponding values into the quadratic formula and calculate.
u = (-(-29) ± √((-29)^2 - 4(1)(100))) / (2(1))
= (29 ± √(841 - 400)) / 2
= (29 ± √441) / 2
= (29 ± 21) / 2
So, the possible values of u are (29 + 21) / 2 = 25 and (29 - 21) / 2 = 4.
Step 6: Solve for x. Now substitute the values of u back into the original substitution equation: u = x^2.
For u = 25:
25 = x^2
x = ±√25
x = ±5
For u = 4:
4 = x^2
x = ±√4
x = ±2
Therefore, the x-intercepts of the equation x^4 - 29x^2 + 100 are x = -5, -2, 2, and 5.