if one zero of the quadratic polynomial 4xsquare + 5x+ gama be reciprocal of another zero, find the value of gama.
4x²+5x+γ=4(x-k)(x-1/k)
=4x²-4(k+1/k)+1
Equate coefficients of terms:
for x²: 4=4
for x: -4(k+1/k)=5
for constant term: γ=1
So the equation is:
4x²+5x+1=0
(4x+1)(x+1)=0
let the roots be p and 1/p
We know that for ax^2 + bx + c = 0
the sum of the roots = -b/a
product of roots = c/a
so in our case, product of the roots = gamma/4
but the product of roots = p(1/p) = 1
so
gamma/4 = 1
gamma = 4
To find the value of "gama," we need to use the relationship between the zeros of the quadratic polynomial.
Let's assume the zeros of the quadratic polynomial 4x^2 + 5x + gama are "a" and "1/a," where a is a nonzero constant.
The sum of the zeros of a quadratic polynomial is given by the formula:
Sum of Zeros = -Coefficient of x / Coefficient of x^2
In this case, the sum of the zeros is:
a + 1/a = -5/4
To simplify the equation, we can multiply both sides by "a":
a^2 + 1 = (-5/4)a
Bringing all terms to one side, we have:
a^2 + (5/4)a + 1 = 0
This equation is in quadratic form. Now, we can solve it using the quadratic formula:
a = (-b ± √(b^2 - 4ac)) / 2a
Here, a = 1, b = (5/4), and c = 1.
Substituting the values into the quadratic formula, we get:
a = (-(5/4) ± √((5/4)^2 - 4(1)(1))) / 2(1)
Simplifying further:
a = (-5 ± √(25/16 - 4)) / 2
a = (-5 ± √(25/16 - 64/16)) / 2
a = (-5 ± √(-39/16)) / 2
Since the discriminant (-39/16) is negative, the equation has no real solutions. Therefore, no value of "gama" can satisfy the given conditions.