Determine the values of k for which the function f(x)= 4x^2-3x+2kx+1 has two zeros , check these values in the original equation
To find the values of k for which the function has two zeros, we need to set the discriminant of the quadratic equation equal to zero. The discriminant (D) is given by:
D = b^2 - 4ac
For our equation f(x) = 4x^2 - 3x + 2kx + 1, the coefficients are a = 4, b = -3 + 2k, and c = 1.
Setting the discriminant equal to zero:
0 = (-3 + 2k)^2 - 4 * 4 * 1
Simplifying and solving for k:
0 = 9 - 6k + 4k^2 - 16
0 = 4k^2 - 6k - 7
Using the quadratic formula to solve for k:
k = (-b ± √(b^2 - 4ac)) / 2a
k = (-(-6) ± √((-6)^2 - 4*4*(-7))) / (2*4)
k = (6 ± √(36 + 112)) / 8
k = (6 ± √148) / 8
Simplifying further:
k = (6 ± 2√37) / 8
Now we have two possible values for k:
k1 = (6 + 2√37) / 8
k2 = (6 - 2√37) / 8
We can substitute these values of k back into the original equation f(x) = 4x^2 - 3x + 2kx + 1 to check if they produce two zeros.
Checking k1:
Let k = (6 + 2√37) / 8
f(x) = 4x^2 - 3x + (2 * (6 + 2√37) / 8) * x + 1
Simplifying:
f(x) = 4x^2 - 3x + (3 + √37) / 4) * x + 1
To find the zeros, we set f(x) equal to zero:
0 = 4x^2 - 3x + (3 + √37) / 4) * x + 1
Checking k2:
Let k = (6 - 2√37) / 8
f(x) = 4x^2 - 3x + (2 * (6 - 2√37) / 8) * x + 1
Simplifying:
f(x) = 4x^2 - 3x + (3 - √37) / 4) * x + 1
To find the zeros, we set f(x) equal to zero:
0 = 4x^2 - 3x + (3 - √37) / 4) * x + 1
Thus, we have found the values of k for which the function has two zeros, and we have checked these values in the original equation.
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 equation, which is given by the formula b^2 - 4ac. In this case, the equation is in the form ax^2 + bx + c, where a = 4, b = (-3 + 2k), and c = 1.
For the equation to have two zeros, the discriminant must be greater than zero. This condition can be expressed as:
(-3 + 2k)^2 - 4(4)(1) > 0
Expanding and simplifying this inequality gives:
9 - 6k + 4k^2 - 16 > 0
Rearranging the terms gives:
4k^2 - 6k - 7 > 0
Now we need to solve this quadratic inequality for k. We can use various methods such as factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
k = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 4, b = -6, and c = -7. Substituting these values into the formula gives:
k = (-(-6) ± √((-6)^2 - 4(4)(-7))) / (2(4))
Simplifying further:
k = (6 ± √(36 + 112)) / 8
k = (6 ± √148) / 8
k = (6 ± 2√37) / 8
Therefore, the two values of k for which the function f(x) has two zeros are:
k1 = (6 + 2√37) / 8
k2 = (6 - 2√37) / 8
Now, to check these values in the original equation f(x), substitute them back into the equation and see if it has two zeros.
For k = (6 + 2√37) / 8:
f(x) for this value of k is:
f(x) = 4x^2 - 3x + 2((6 + 2√37)x) + 1
And for k = (6 - 2√37) / 8:
f(x) for this value of k is:
f(x) = 4x^2 - 3x + 2((6 - 2√37)x) + 1
Evaluate the functions for different values of x to determine if they have two zeros.