Let's assume that the zeros of the polynomial 6x^2 + 37x - (k-2) are a and b. We know that a and b are reciprocals of each other. Therefore, we can write:
a * b = 1
Also, we know that the sum of the zeros of a quadratic polynomial is given by -b/a. Therefore, we can write:
a + b = -37/6
We can use these two equations to find the values of a and b. Solving the first equation for b, we get:
b = 1/a
Substituting this in the second equation, we get:
a + (1/a) = -37/6
Multiplying both sides by a, we get:
a^2 + 1 = (-37/6) * a
Multiplying both sides by 6, we get:
6a^2 + 6 = -37a
Rewriting the given polynomial in standard form by adding (k-2) to both sides, we get:
6x^2 + 37x - (k-2) = 0
Comparing the coefficients of x^2 and x, we get:
a + b = -37/6 => a + 1/a = -37/6
a * b = 1 => ab = 1
Substituting b = 1/a in the second equation, we get:
a * (1/a) = 1 => a^2 = 1 => a = ±1
Substituting these values of a in the equation a + 1/a = -37/6, we get:
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Step-by-step explanation:
!
Let's assume that the zeros of the polynomial 6x^2 + 37x - (k-2) are a and b. We know that a and b are reciprocals of each other. Therefore, we can write:
a * b = 1
Also, we know that the sum of the zeros of a quadratic polynomial is given by -b/a. Therefore, we can write:
a + b = -37/6
We can use these two equations to find the values of a and b. Solving the first equation for b, we get:
b = 1/a
Substituting this in the second equation, we get:
a + (1/a) = -37/6
Multiplying both sides by a, we get:
a^2 + 1 = (-37/6) * a
Multiplying both sides by 6, we get:
6a^2 + 6 = -37a
Rewriting the given polynomial in standard form by adding (k-2) to both sides, we get:
6x^2 + 37x - (k-2) = 0
Comparing the coefficients of x^2 and x, we get:
a + b = -37/6 => a + 1/a = -37/6
a * b = 1 => ab = 1
Substituting b = 1/a in the second equation, we get:
a * (1/a) = 1 => a^2 = 1 => a = ±1
Substituting these values of a in the equation a + 1/a = -37/6, we get:
a = 1 => 1 + 1/1 = -37/6 (not possible)
a = -1 => -1 + 1/-1 = -37/6 => -2 = -37/6 => 37/6 = 2
Therefore, the value of k is given by the constant term in the polynomial:
k - 2 = -ab = -1 * (-1) = 1
k =3
Answer:
The value of k is 2668/111.
Step-by-step explanation:
Let the zeros of the polynomial be a and 1/a.
Using the sum and product of roots formula, we have:
a + (1/a) = -37/6 ---(1)
a(1/a) = -(k-2)/6 ---(2)
Multiplying equation (1) by a, we get:
a^2 + 1 = (-37/6)a
Multiplying equation (1) by 1/a, we get:
1 + a^2 = (-37/6)(1/a)
Adding the above two equations, we get:
a^2 + 1 + 1 + a^2 = (-37/6)a + (-37/6)(1/a)
Simplifying, we get:
2a^2 + 2 = (-37/6)(a + 1/a)
Substituting (1) in the above equation, we get:
2a^2 + 2 = (-37/6)(-37/6)
Simplifying, we get:
2a^2 + 2 = 1369/36
Multiplying both sides by 18, we get:
36a^2 + 36 = 1369
Simplifying, we get:
36a^2 = 1333
Dividing both sides by 36, we get:
a^2 = 1333/36
Taking the square root, we get:
a = ±√(1333)/6
Since one zero is the reciprocal of the other, we must take the positive value of a.
Now, substituting (1) with a = √(1333)/6, we get:
√(1333)/6 + 6/√(1333) = -37/6
Multiplying both sides by √(1333)/6, we get:
1333/36 - 36/1333 = -37/6 * √(1333)/6
Simplifying, we get:
(k-2) = 2666/111
Therefore, k = 2668/111.
Hence, the value of k is 2668/111.
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