Question

If are the zeroes of the quadratic polynomial f(x)=3x2 -4x+1, Find a quadratic polynomial whose zeroes are 2/ 2/ ?​

Answer 1

Answer:

The required quadratic polynomial is 2x²+3x-1

Step-by-step explanation:

If \alphaα and \betaβ are the zeroes of quadratic polynomial

f(x)=x^2-3x-2f(x)=x2−3x−2 then

\alpha+\beta=-\frac{\text{Coefficient of x}}{\text{Coefficient of } x^2}α+β=−Coefficient of x2Coefficient of x

\implies \alpha+\beta=-(\frac{-3}{1})=3⟹α+β=−(1−3)=3

Also,

\alpha\beta=\frac{\text{Constant Term}}{\text{Coefficient of }x^2}αβ=Coefficient of x2Constant Term

\implies \alpha\beta=\frac{-2}{1}=-2⟹αβ=1−2=−2

We have to find the quadratic polynomial whose zeroes are \alpha+\frac{1}{\beta}α+β1 and \beta+\frac{1}{\alpha}β+α1

Sum of zeroes

=\alpha+\frac{1}{\beta}+\beta+\frac{1}{\alpha}=α+β1+β+α1

=(\alpha+\beta)+(\frac{1}{\alpha}+\frac{1}{\beta})=(α+β)+(α1+β1)

=3+\frac{\alpha+\beta}{\alpha\beta}=3+αβα+β

=3+\frac{3}{-2}=3+−23

=-\frac{3}{2}=−23

Product of zeroes

=(\alpha+\frac{1}{\beta})\times(\beta+\frac{1}{\alpha})=(α+β1)×(β+α1)

=\alpha\beta+1+1+\frac{1}{\alpha\beta}=αβ+1+1+αβ1

=-2+2+\frac{1}{-2}=−2+2+−21

=-\frac{1}{2}=−21

We know that quadratic polynomial whose zeroes are \alphaα and \betaβ is given by

x^2-(\alpha+\beta)x+\alpha\betax2−(α+β)x+αβ

Therefore, the quadratic polynomial is

x^2-(-\frac{3}{2})x+(-\frac{1}{2})x2−(−23)x+(−21)

or, x^2+\frac{3}{2}x-\frac{1}{2}x2+23x−21

or, 2x^2+3x-12x2+3x−1

Step-by-step explanation:

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