Question

find the quadratic polynomial whose zeros are 1 /a and 1/2a​

Answer 1

AnswEr :

The required polynomial is 2a²x² - 3ax + 1

Given that,

1/a and 1/2a are the zeros of the required polynomial

Let m and n be the zeros of the polynomial

• The polynomial would be of the form K[x² - (m + n) + mn]

Here,

Sum of Zeros

$\sf \: m + n = \dfrac{1}{a} + \dfrac{1}{2a} \\ \\ \longrightarrow \: \sf \: m + n = \dfrac{3a}{2 {a}^{2} }$

Product of Zeros

$\sf \: mn = \dfrac{1}{a} \times \dfrac{1}{2a} \\ \\ \longrightarrow \: \sf \: mn = \dfrac{1}{2 {a}^{2} }$

Let the polynomial be p(x)

Thus,

$\sf \: p(x) = k\bigg( {x}^{2} - (m + n)x + mn) \bigg)$

If k = 2a²,then

$\longrightarrow \boxed{ \boxed{ \sf \: p(x) = 2 {a}^{2} {x}^{2} - 3ax + 1}}$

Answer 2

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QUESTION :

find the quadratic polynomial whose zeros are 1 /a and 1/2a.

SOLUTION :

We know that a Quadratic Polynomial can be expressed in the following from :

X^2 - { Sum of Zeroes } + { Product of Zeroes }

Zeroes are 1 / a and 1 / 2a.

Sum of Zeroes = 1 / a + 1 / 2a

=> { 1 / a } [ 1 + 1/2 ]

=> 3 / 2a

Product Of Zeroes :

1 / a × 1 / 2a

=> 1 / 2 a^2

Hence The quadratic equation becomes :

X^2 - 3 / 2a + 1 /2a^2 = 0

Muliplying by 2a^2

=> 2a^2X^2 - 3a + 1 = 0

Answer : 2a^2X^2 - 3a + 1 = 0