if zeroes of p(x)=5x^2-7x+k are reciprocal of each other then find value of k
using the quadratic formula,
x = [7±√(49-20k)]/10
so,
(7+√(49-20k))/10 * (7-√(49-20k))/10 = 1
49-(49-20k) = 100
20k = 100
k = 5
If the zeroes of a quadratic polynomial are reciprocal of each other, it means that if one zero is 'a', then the other zero is '1/a'.
Let's assume that 'a' is one of the zeroes. Then, the other zero will be '1/a'.
Now, the sum of the zeroes of a quadratic polynomial is given by the formula:
Sum of zeroes = -coefficient of x / coefficient of x^2
In this case, the sum of the zeroes is:
a + (1/a) = (-(-7)) / 5
To simplify the equation, we can multiply both sides by 'a' to eliminate the denominator:
a^2 + 1 = 7/5 * a
Multiplying both sides by 5:
5a^2 + 5 = 7a
Rearranging the equation:
5a^2 - 7a + 5 = 0
Comparing this equation with the standard form of a quadratic equation ax^2 + bx + c = 0, we can see that:
a = 5, b = -7, c = 5
Now, we can use the quadratic formula to solve for 'a':
a = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values:
a = (-( -7) ± √((-7)^2 - 4 * 5 * 5)) / (2 * 5)
Simplifying:
a = (7 ± √(49 - 100)) / 10
Since the discriminant (b^2 - 4ac) is negative, the roots will be complex numbers. Therefore, there are no real values of 'a' that satisfy the given condition. Hence, there is no specific value of 'k' for which the zeroes of p(x) = 5x^2 - 7x + k are reciprocal of each other.
To find the value of k, we first need to understand what it means for the zeroes of a quadratic polynomial to be reciprocal of each other.
The zeroes of a polynomial are the values of x for which the polynomial equals zero. In this case, the quadratic polynomial p(x) = 5x^2 - 7x + k has two zeroes.
If the zeroes of the polynomial are reciprocal of each other, it means that if one zero is denoted by a, then the other zero is 1/a. In equation form, this means:
a * (1/a) = 1
Now, let's find the zeroes of the given quadratic polynomial:
To find the zeroes of p(x), we need to solve the equation 5x^2 - 7x + k = 0.
Using the quadratic formula, x = (-b ± sqrt(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation (in this case, a = 5, b = -7, c = k):
x = (-(-7) ± sqrt((-7)^2 - 4 * 5 * k)) / (2 * 5)
= (7 ± sqrt(49 - 20k)) / 10
So the zeroes of the quadratic polynomial are:
x1 = (7 + sqrt(49 - 20k)) / 10
x2 = (7 - sqrt(49 - 20k)) / 10
Since the zeroes are reciprocal of each other, we have:
x1 * x2 = 1
Substituting the values of x1 and x2, we get:
[(7 + sqrt(49 - 20k)) / 10] * [(7 - sqrt(49 - 20k)) / 10] = 1
Expanding and simplifying the equation, we have:
(49 - (sqrt(49 - 20k))^2) / 100 = 1
Now, let's solve for k:
49 - (sqrt(49 - 20k))^2 = 100
- (sqrt(49 - 20k))^2 = 100 - 49
- (sqrt(49 - 20k))^2 = 51
(sqrt(49 - 20k))^2 = -51 (Note: Since we are dealing with real numbers, this equation has no solution.)
Therefore, there is no value of k that satisfies the condition that the zeroes of the polynomial p(x) = 5x^2 - 7x + k are reciprocal of each other.