Question

If Alfa and beta are the zeroes of the polynomial 6y^2-7y+2,find a quadratic polinomial whose zeroes are 1/alfaand1/beta

Answer 1

Answer:

α and β are the zeroes of the polynomial 6y²-7y+2

∴α+β=-b/a=7/6

 αβ=c/a=1/3

for the new polynomial,

sum of zeroes=1/α+1/β=α+β/αβ=7/6*3=7/2

product of zeroes=1/α*1/β=1/αβ=3

∴new polynomial=y²-(sum of zeroes)y+(product of zeroes)

                            =y²-7/2y+3

                            =2y²-7y+6

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Step-by-step explanation:

Answer 2

Answer :

The required quadratic polynomial is : 2x² - 7x + 6

Given :

The quadratic polynomial is :

• 6y² -7y + 2
• α and β are the zeroes of the given polynomial

Formulae to be used :

If α and β are the zeroes of a polynomial then the following relationships between zeroes and coefficient is given by :

$\sf \star \: \: \alpha + \beta = -\dfrac{coefficient \: \: of \: \: x}{Coefficient \: \: of \: \: x^{2}} \\\\ \sf \star \: \: \alpha\beta = \dfrac{Constant \: \: term}{Coefficient\: \: of \: \: x^{2}}$

And the expression for a polynomial can also be given by :

$\sf \star \: \: x^{2} - ( \alpha + \beta) + \alpha \beta$

Solution :

From the given equation :

Sum of the roots

$\sf \alpha + \beta = -\dfrac{-7}{6} \\\\ \sf \implies \alpha + \beta = \dfrac{7}{6} .......... (1)$

and product of the roots :

$\sf \implies \alpha \beta = \dfrac{2}{6}\\\\ \sf \implies \alpha \beta = \dfrac{1}{3}........ (2)$

Dividing (1) by (2) we have :

$\sf \implies \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{\dfrac{7}{6}}{\dfrac{1}{3}} \\\\ \sf \implies \dfrac{\alpha}{\alpha\beta} + \dfrac{\beta}{\alpha\beta} = \dfrac{7\times 3 }{6} \\\\ \sf \implies \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{7}{2}$

Now the required quadratic polynomial is :

$\sf x^{2} - \dfrac{7}{2}x + \dfrac{1}{\dfrac{1}{3}} \\\\ \sf = x^{2} - \dfrac{7}{2}x + 3 \\\\ \sf \dagger \: \: Multiplying \: by \: 2 \: we \: get \\\\ \sf = 2 \{x^{2} - \dfrac{7}{2}+ 3 \} \\\\ \sf = 2x^{2} - 7x + 6$