If the quadratic equation, which is in the form
ax^{2} + 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 :
2x² + 7/2x + 3/4 = 0
Solution :
2x² + 7/2x + 3/4 = 0
8x² + 14x + 3 = 0
8x² + 2x + 12x + 3 = 0
2x(4x + 1) + 3(4x + 1) = 0
(2x + 3) (4x + 1) = 0
2x + 3 = 0 4x + 1 = 0
x = -3/2 x = -1/4
So, α = -3/2 and β = -1/4
Verifying relationship :
2x² + 7/2x + 3/4 = 0
ax² + bx + c = 0
a = 2, b = 7/2, c = 3/4
Sum of zeros : α + β = -b/a (-3/2) + (-1/4) = -(7/2)/2 -7/4 = -7/4 |
Product of zeros : αβ = c/a (-3/2)(-1/4) = (3/4)/2 3/8 = 3/8 |
Verified.
Problem 2 :
4x² + 5√2x - 3 = 0
Solution :
First, let us find the zeros of the polynomial.
4x² + 5√2x - 3 = 0
= 4x² + 6√2x - √2x - 3
= 2√2x (√2x + 3) - 1(√2x + 3)
= (2√2x - 1) (√2x + 3)
2√2x - 1 = 0 √2x + 3 = 0
x = 1/2√2 x = -3/√2
So, α = 1/2√2 and β = -3/√2
Verifying the relationship :
4x² + 5√2x - 3 = 0
ax² + bx + c = 0
a = 4, b = 5√2, c = -3
Sum of zeros : α + β = -b/a 1/2√2 + (-3/√2) = -(5√2/4) -5√2/4 = -5√2/4 |
Product of zeros : αβ = c/a (1/2√2)(-3/√2) = -3/4 -3/4 = -3/4 |
Verified.
Problem 3 :
2s² - (1 + 2√2)s + √2 = 0
Solution :
First, let us find the zeros of the polynomial.
2s² - (1 + 2√2)s + √2 = 0
= 2s² - s - 2√2s + √2
= s(2s - 1) - √2 (2s - 1)
= (2s - 1) (s - √2)
2s - 1 = 0 s - √2 = 0
s = 1/2 s = √2
So, α = 1/2 and β = √2
Verifying the relationship :
2s² - (1 + 2√2)s + √2 = 0
ax² + bx + c = 0
a = 2, b = -(1 + 2√2), c = √2
Sum of zeros : α + β = -b/a 1/2 + √2 = -(-(1 + 2√2))/2 1 + 2√2/2 = 1 + 2√2/2 |
Product of zeros : αβ = c/a (1/2)(√2) = √2/2 √2/2 = √2/2 1/√2 = 1/√2 |
Verified.
Problem 4 :
v² + 4√3v - 15 = 0
Solution :
v² + 4√3v - 15 = 0
= v² + 5√3v - √3v - 15
= v(v + 5√3) - √3(v + 5√3)
= (v + 5√3) (v - √3)
v + 5√3 = 0 v - √3 = 0
v = -5√3 v = √3
So, α = -5√3 and β = √3
Verifying the relationship :
v² + 4√3v - 15 = 0
ax² + bx + c = 0
a = 1, b = 4√3, c = -15
Sum of zeros : α + β = -b/a (-5√3) + √3 = -(4√3)/1 -4√3 = -4√3 |
Product of zeros : αβ = c/a (-5√3) (√3) = -15/1 -15 = -15 |
Verified.
Problem 5 :
y² + 3/2√5y - 5 = 0
Solution :
y² + 3/2√5y - 5 = 0
2y² + 3√5y - 10 = 0
2y² + 4√5y - √5y - 10 = 0
2y(y + 2√5) - √5(y + 2√5)
(y + 2√5) (2y - √5)
y + 2√5 = 0 2y - √5 = 0
y = -2√5 y = √5/2
So, α = -2√5 and β = √5/2
Verifying the relationship :
v² + 4√3v - 15 = 0
ax² + bx + c = 0
a = 2, b = 3√5, c = -10
Sum of zeros : α + β = -b/a (-2√5) + √5/2 = -(3√5)/2 -3√5/2 = -3√5/2 |
Product of zeros : αβ = c/a (-2√5) (√5/2) = -10/2 -5 = -5 |
Verified.
Problem 6 :
7y² - 11/3y - 2/3 = 0
Solution :
7y² - 11/3y - 2/3 = 0
(21y² - 11y - 2)/3 = 0
21y² - 11y - 2 = 0
21y² - 14y + 3y - 2 = 0
7y(3y - 2) + 1(3y - 2) = 0
(3y - 2) (7y + 1) = 0
3y - 2 = 0 7y + 1 = 0
y = 2/3 y = -1/7
So, α = 2/3 and β = -1/7
Verifying relationship :
7y² - 11/3y - 2/3 = 0
ax² + bx + c = 0
a = 7, b = -11/3, c = -2/3
Sum of zeros : α + β = -b/a 2/3 + (-1/7) = -(-11/3)/7 11/21 = 11/21 |
Product of zeros : αβ = c/a (2/3) (-1/7) = (-2/3)/7 -2/21 = -2/21 |
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