Determine the values of k for which the function f(x)=4x^2-3x+2kx+1 has two zeros. Check these values in the orginal equation.
If there are two distinct roots, the determinant is positive:
(-3+2k)^2 - 4(4)(1) > 0
9 - 12k + 4k^2 - 16 > 0
4k^2 - 12k -7 > 0
(2k+1)(2k-7) > 0
Either both factors are positive or both are negative
k < -1/2 or k > 7/2
where does 12k come from
To determine the values of k for which the function f(x) = 4x^2 - 3x + 2kx + 1 has two zeros, we need to find the discriminant of the quadratic equation formed by setting f(x) equal to zero.
The discriminant, denoted as Δ, is calculated using the formula Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
In this case, the coefficients are a = 4, b = -3 + 2k, and c = 1. Substituting these values into the discriminant formula, we have Δ = (-3 + 2k)^2 - 4(4)(1).
Next, we set the discriminant to be greater than zero, which indicates that there are two distinct real roots. Therefore, we have:
(-3 + 2k)^2 - 4(4)(1) > 0
Expanding and simplifying the equation, we obtain:
9 - 6k + 4k^2 - 16 > 0
Rearranging the terms, we have:
4k^2 - 6k - 7 > 0
Now, we need to solve this inequality to determine the values of k that satisfy it.
To do this, we can factorize the quadratic expression or use the quadratic formula. In this case, let's use the quadratic formula:
k = (-b ± √(b^2 - 4ac)) / (2a)
Substituting a = 4, b = -6, and c = -7 into the quadratic formula, we obtain:
k = (-(-6) ± √((-6)^2 - 4(4)(-7))) / (2(4))
k = (6 ± √(36 + 112)) / 8
k = (6 ± √148) / 8
k = (6 ± 2√37) / 8
k = (3 ± √37) / 4
Therefore, the values of k that satisfy the inequality and give two distinct real roots for the function are k = (3 + √37) / 4 and k = (3 - √37) / 4.
To check these values in the original equation, substitute them back into f(x) = 4x^2 - 3x + 2kx + 1 and see if the equation holds true for x values that satisfy the original problem.