Question

if alpha and beta are two zeros of a quadratic equation X² + 2x- 35 find the quadratic polynomial whose zeros are 1/alpha and 1/beta​

Answer 1

Answer :-

The quadratic polynomial whose zeroes are 1/α and 1/β is 35x² - 2x - 1

Solution :-

x² + 2x - 35

To find zeroes equate it to 0

⇒ x² + 2x - 35 = 0

⇒ x² + 7x - 5x - 35 = 0

⇒ x(x + 7) - 5(x + 7) = 0

⇒ (x - 5)(x + 7) = 0

⇒ x - 5 = 0 or x + 7 = 0

⇒ x = 5 or x = - 7

5 and - 7 are zeroes of a quadratic polynomial x² + 2x - 35.

Given α and β are zerores of a quadratic polynomial x² + 2x - 35.

So

• α = 5

• β = - 7

Quadratic polynomial whose zeroes are 1/α and 1/β

α = 5

⇒ 1/α = 1/5

β = - 7

⇒ 1/β = - 1/7

Zeroes are 1/5 and - 1/7

So

α = 1/5

β = - 1/7

Now find sum of zeroes and product of zeroes

Sum of zeroes = α + β

= 1/5 + ( - 1/7)

= 1/5 - 1/7

= (7 - 5)/35

= 2/35

Product of zeroes = αβ

= 1/5(- 1/7)

= - 1/35

Quadratic polynomial ax² + bx + c = k[x² - x(α + β) + αβ]

(Where k ≠ 0)

Here

• α + β = 2/35

• αβ = - 1/35

By substituting the values

= k[x² - x(2/35) + (-1/35)]

= k(x² - 2x/35 - 1/35)

= k[(35x² - 2x - 1)/35]

When k = 35

= 35[(35x² - 2x - 1)/35]

= 35x² - 2x - 1

Therefore the quadratic polynomial whose zeroes are 1/α and 1/β is 35x² - 2x - 1

Answer 2

Answer :

The quadratic polynomial whose zeroes are 1/α and 1/β is 35x² - 2x - 1

Step-by-step explanation :

Given that :

α and β are two zeros of a quadratic equation x² + 2x- 35.

To Find :

The quadratic polynomial whose zeros are 1/α & 1/β​.

Solution :

$\bigstar\;\textbf{\underline{\underline{Consider as -}}}}$

The zeroes of p(x) x² + 2x- 35 as - α & β

So,

$\bigstar\;\boxed{\sf Product\;of\;the\;zeroes = \alpha.\beta}}\\ \\\bf{\underline{\underline{Values,}}}\\ \sf \\\implies \dfrac{c}{a} = \dfrac{-35}{1} = -35\\ \\ \\\star\;\boxed{\sf \alpha.\beta=-35.}}$

A Polynomial with zeroes :

1/α and 1/β.

So,

$\therefore\;\sf Sum = \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\beta +\alpha}{\alpha \beta} = \dfrac{\alpha+\beta}{\alpha \beta}\\ \\ \\\implies \dfrac{-2}{35} = \dfrac{2}{35}$

$\sf Product = \left(\dfrac{1}{\alpha}\right),\left(\dfrac{1}{\beta}\right)\\ \\ \\\implies \dfrac{1}{\alpha \beta} = \dfrac{1}{-35}$

★ New Polynomial :

$\sf \\ \\\implies x^2 - (\alpha + \beta)n\;+ \alpha \beta\quad \quad [General\;formula\;for\;polynomial]\\ \\\implies x^2 \left[\dfrac{2}{35}\;x+\left[\dfrac{1}{-35}\right]$

$\sf \\ \\\implies x^2 - \dfrac{2}{35},x-\dfrac{1}{35}\\ \\ \\\implies 35 \left(x^2- \dfrac{2}{35},x-\dfrac{1}{35}\right)\\ \\ \\\implies 35x^2 - 2x - 1$

★ Hence,

The quadratic polynomial whose zeroes are 1/α and 1/β is 35x² - 2x - 1