VERIFYING THE RELATION BETWEEN ZEROES AND COEFFICIENTS

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 :

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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