RELATIONSHIP BETWEEN ZEROS AND COEFFICIENTS OF A POLYNOMIAL

Problem 1 :

If α and β are zeros of p(x) = x² + x - 1, then find 1/α+ 1/β?

Solution :

p(x) = x² + x - 1

Here a = 1, b = 1 and c = -1

Finding 1/α+ 1/β :

Since the denominators are not same,

1/α + 1/β = (β + α)/αβ

1/α + 1/β = (α+β)/αβ -----(1)

Sum of zeroes :

α + β = -1/1 ==> -1

Product of zeroes :

α β = -1/1 ==> -1

Applying these values in (1), we get

= -1/(-1)

1/α + 1/β = 1

Problem 2 :

If one of the zeros of the cubic polynomial x³ + ax² + bx + c is -1, then what will be the product of the other two zeros?

Solution :

Any cubic polynomial x³ + ax² + bx + c will have three zeroes.

The zeroes are α, β and δ

Let δ = -1

Product of the roots 

α β δ = -c/1

α β δ = -c

Applying the value of δ, we get

α β (-1) = -c

α β = c ----(1)

Since -1 is one of the zero of p(x) = x³ + ax² + bx + c 

p(-1) = (-1)³ + a(-1)² + b(-1) + c

0 = -1 + a - b + c

c = 1 - a + b

Applying the value of c, we get

α β = 1 - a + b

Problem 3 :

If α, β, γ be zeros of the polynomial 6x³ + 3x² -5x+1, then find the value of α^-1 + β^-1 + γ^-1?

Solution :

p(x) = 6x³ + 3x² - 5x + 1

a = 6, b = 3, c = -5 and d = 1

α^-1 + β^-1 + γ^-1 = 1/α + 1/β + 1/γ

α^-1 + β^-1 + γ^-1 = (βγ + αγ + βα) / αβγ ---(1)

α β + β γ + γ α = -5/6

αβγ = -1/6

Applying the values in (1), we get

α^-1 + β^-1 + γ^-1 = (-5/6)/(-1/6)

= (-5/6) x (-6/1)

= 5

Problem 4 :

If α, β are zeros of the quadratic polynomial

f(x) = 2x² + 11x + 5

find

a) α⁴ + β⁴      b)1/α +1/β -2αβ

Solution :

a) α⁴ + β⁴ = (α²)² + (β²)²

= (α² + β²)² - 2(α² β²)

= (α² + β²)² - 2(α β)²

= [(α + β)² - 2α β]² - 2(α β)²  ---(1)

From f(x),  α + β = -11/2, αβ = 5/2

Applying the above values in (1), we get

 = [(-11/2)² - 2(5/2)]² - 2(5/2)²

 = [(121/4) - 5]² - (25/2)

 = [(121 - 20)/4]² - (25/2)

 = [101/4]² - (25/2)

= (10201 - 200) / 16

= 10001/16

b)1/α +1/β -2αβ

= [(β + α)/αβ] - 2αβ

= [(α + β)/αβ] - 2αβ ---(2)

From f(x), α + β = -11/2, αβ = 5/2

Applying the above values in (2), we get

= [(-11/2)/(5/2)] - 2(5/2)

= -11/5 - 5

= (-11 - 25) /5

= -36/5

Problem 5 :

If the zeros of the polynomial f(x) = x³ – 3x² - 6x + 8 are of the form a-b, a, a+b, then find all the zeros

Solution :

f(x) = x³ – 3x² - 6x + 8 

Since the zeroes are a-b, a, a+b,

Sum of zeroes :

a - b + a + a + b = -(-3)/1

3a = 3

a = 1

Product of zeroes :

(a - b) a (a + b) = -8/1

Applying the value of a, we get

(1 - b) 1 (1 + b) = -8

1 - b² = -8

b² = 1 + 8

b² = 9

b = 3 or -3

So, the required zeroes are -2, 1 and 4.

Problem 6 :

If α, β are the two zeros of the polynomial

f(y) = y² - 8y + a and α² + β² = 40

find the value of ‘a’?

Solution :