Question

Form a polynomial whose zeroes are reciprocal of zeroes of 6x² - 7x + 2. *​

Answer 1

Given to find the polynomial whose zeros are reciprocal of 6x² - 7x + 2 :-

SOLUTION:-

Substitute in place of x, 1/x which gives the polynomial whose zeros are reciprocal

$6x {}^{2} - 7x + 2 = 0$

$6 \bigg( \dfrac{1}{x} \bigg) {}^{2} - 7 \bigg( \dfrac{ 1}{x} \bigg) + 2 = 0$

$6 \bigg(\dfrac{1}{x {}^{2} } \bigg) - \dfrac{7}{x} + 2 = 0$

$\dfrac{6}{x {}^{2} } - \dfrac{7}{x} + 2 = 0$

Take L.C.M to the denominators

$\dfrac{6 - 7x + 2x {}^{2} }{x {}^{2} } = 0$

$6 - 7x + 2x {}^{2} = 0(x {}^{2} )$

$6 - 7x + 2x {}^{2} = 0$

$2x {}^{2} - 7x + 6 = 0$

This is the required polynomial whose roots are reciprocal to each other

Know more :-

Transformation of equations:-

If ${ \alpha, \beta}$ are the roots of Quadratic equation f(x) ax² + bx + c = 0 then the Quadratic equation whose roots are

☆ The opposite signs ${ -\alpha, - \beta}$ of the roots of f(x) = 0 is f(-x) =0

☆ If it is increased by k ${ \alpha + k, \beta + k}$ more than that of f(x ) =0 is f(x -k ) = 0

☆ If it is decreased by k ${ \alpha - k, \beta - k}$ more than that of f(x ) =0 is f(x +k ) = 0

☆ Multiplied by k ${ k\alpha, k \beta}$ of the roots of f(x) = 0 is f(x/k) = 0

Answer 2

Given :- Form a polynomial whose zeroes are reciprocal of zeroes of 6x² - 7x + 2. ?

Answer :-

putting the given equation to zero and finding the roots first .

→ 6x² - 7x + 2 = 0

→ 6x² - 3x - 4x + 2 = 0

→ 3x(2x - 1) - 2(2x - 1) = 0

→ (2x - 1)(3x - 2) = 0

→ x = (1/2) and (2/3) = zeroes of the equation .

now, reciprocal of these zeroes will be,

• (1/2) reciprocal = 1/(1/2) = 2
• (2/3) reciprocal = 1/(2/3) = 3/2

then, the required polynomial will be,

→ x² - (sum of zeroes)x + product of zeroes

→ x² - (2 + 3/2)x + 2 * (3/2)

→ x² - {(4 + 3)/2}x + 3

→ x² - (7x/2) + 3

→ (2x² - 7x + 6)/2

→ 2x² - 7x + 6 (Ans.)

Learn more :-

JEE mains Question :-

https://brainly.in/question/22246812

. Find all the zeroes of the polynomial x4

– 5x3 + 2x2+10x-8, if two of its zeroes are 4 and 1.

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