If the quadratic equation, which is in the form
ax2 + bx + c = 0
Sum and product of zeroes can be found using the formulas given below.
Sum of the zeros :
α + β = - b/a
Product of zeros :
α × β = c / a
Find the zeros of the following polynomials by factorization method and verify the relations between the zeros and the coefficients of the polynomials:
Problem 1 :
4x² - 3x - 1 = 0
Solution :
4x² - 3x - 1 = 0
4x² - 4x + x - 1 = 0
4x(x - 1) + 1(x - 1) = 0
(4x + 1) (x - 1) = 0
4x + 1 = 0 x - 1 = 0
x = -1/4 x = 1
So, α = -1/4 and β = 1
Verifying relationship :
4x² - 3x - 1 = 0
ax² + bx + c = 0
a = 4, b = -3, c = -1
Sum of zeros : α + β = -b/a -1/4 + 1 = - (-3)/4 3/4 = 3/4 |
Product of zeros : αβ = c/a (-1/4)(1) = -1/4 -1/4 = -1/4 |
Verified.
Problem 2 :
3x² + 4x - 4 = 0
Solution :
First, let us find the zeros of the polynomial.
3x² + 4x - 4 = 0
3x² + 6x - 2x - 4 = 0
3x(x + 2) - 2(x + 2) = 0
(3x - 2) (x + 2) = 0
3x - 2 = 0 x + 2 = 0
x = 2/3 x = -2
So, α = 2/3 and β = -2
Verifying relationship :
3x² + 4x - 4 = 0
ax² + bx + c = 0
a = 3, b = 4, c = -4
Sum of zeros : α + β = -b/a 2/3 - 2 = -4/3 -4/3 = -4/3 |
Product of zeros : αβ = c/a (2/3)(-2) = -4/3 -4/3 = -4/3 |
Hence, verified the relations between the zeros and the coefficients of the polynomial.
Problem 3 :
5t² + 12t + 7 = 0
Solution :
First, let us find the zeros of the polynomial.
5t² + 12t + 7 = 0
5t² + 5t + 7t + 7 = 0
5t(t + 1) + 7(t + 1) = 0
(5t + 7) (t + 1) = 0
5t + 7 = 0 t + 1 = 0
t = -7/5 t = -1
So, α = -7/5 and β = -1
Verifying relationship :
5t² + 12t + 7 = 0
ax² + bx + c = 0
a = 5, b = 12, c = 7
Sum of zeros : α + β = -b/a -7/5 - 1 = -12/5 -12/5 = -12/5 |
Product of zeros : αβ = c/a (-7/5)(-1) = 7/5 7/5 = 7/5 |
Verified.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM