Question

The quadratic equation whose roots are the reciprocals of the roots of the Q.E. 3x2 – 20x +
17 = 0 is

Answer 1

$\underline{\underline{\orange{\textbf{Step - By - Step - Explanation : -}}}}$

To Find :

• Quadratic equation whose roots are reciprocal of roots of quadratic equation 3x² - 20x + 17

Solution :

Let α and β are roots of given equation.

3x² - 20x + 17

here

• a = 3
• b = -20
• c = 17

Sum of zeroes = -b/a

α + β = -(-20)/3

α + β = 20/3

Product of zeroes = c/a

αβ = 17/3

Now, we have to find the quadratic equation whoose roots are 1/a and 1/β

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

›› x² - (1/α + 1/β)x + αβ

›› x² - [( β+ α)/αβ]x + αβ

›› x² - [(20/3)/(17/3)]x + 17/3

›› x² - [20/3 × 3/17]x + 17/3

›› x² - 20x/17 + 17/3

›› (51x² - 60x + 289)/51

›› 51x² - 60x + 289

Hence

• Required Quadratic equation is 51x² - 60x + 289

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Answer 2

$\huge\sf\pink{Answer}$

☞ Your answer is 51x²-60x+289

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$\huge\sf\blue{Given}$

✭ P(x) = 3x²-20x+17 = 0

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$\huge\sf\gray{To \:Find}$

◈ A Quadratic polynomial who's roots are the reciprocal of the zeros of the Polynomial p(x)

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$\huge\sf\purple{Steps}$

Assume the roots of the polynomial as α & β which means the new polynomial should have its roots as 1/α & 1/β

On comparing p(x) with ax²+bx+c

◕ a = 3

◕ b = -20

◕ c = 17

So now we know that,

$\underline{\boxed{\sf Sum \ of \ zeros = \dfrac{-b}{a}}}$

Substituting the values,

➝ $\sf \alpha+\beta = \dfrac{-b}{a}$

➝ $\sf \alpha+\beta = \dfrac{-(-20)}{3}$

➝ $\sf \green{\alpha+\beta = \dfrac{20}{3}}$

Now Product of Zeros is given by,

$\underline{\boxed{\sf Product \ of \ zeros = \dfrac{c}{a}}}$

Substituting the values,

➳ $\sf \alpha\beta = \dfrac{c}{a}$

➳ $\sf \red{\alpha\beta = \dfrac{17}{3}}$

Now that we know these the polynomial is given by,

$\underline{\boxed{\sf{x^2- (Sum \ of \ zeros)x+Product \ of \ zeros}}}$

So the zeros of the new polynomial say q(x) are

• 1/α
• 1/β

Substituting the values,

»» $\sf q(x) = x^2-(Sum \ of \ zeros)x+Product of zeros$

»» $\sf x^2-(\dfrac{1}{\alpha}+\dfrac{1}{\beta})x+\dfrac{17}{3}$

»» $\sf x^2-\bigg\lgroup \dfrac{\alpha+\beta}{\alpha\beta}\bigg\rgroup + \dfrac{17}{3}$

»» $\sf x^2-\bigg\lgroup \dfrac{20}{3}\times \dfrac{3}{17} \bigg\rgroup +\dfrac{17}{3}$

»» $\sf x^2- \dfrac{20}{17}x+\dfrac{17}{3}$

[17×3 = 51]

»» $\sf 51\bigg\lgroup x^2- \dfrac{20}{17}x+\dfrac{17}{3}\bigg\rgroup$

»» $\sf \orange{q(x) = 51x^2-60x+289}$

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