Question

if alpha and beta are the zeroes of qp f(x) = x²- 1 find the qp whose zeroes are 2 Alpha divided by beta and 2 beta divided by alpha​

Answer 1

Answer:

GIVEN :

\alphaα and \betaβ are the zeroes of x^2-1x

2

−1

TO FIND :

A quadratic polynomial whose zeroes are \frac{2\alpha}{\beta}

β

2α

and \frac{2\beta}{\alpha}

α

2β

SOLUTION :

Given that \alphaα and \betaβ are the zeroes of x^2-1x

2

−1

x^2-1=0x

2

−1=0

x^2-1^2=0x

2

−1

2

=0

The algebraic identity is given by:

a^2-b^2=(a-b)(a+b)a

2

−b

2

=(a−b)(a+b)

(x-1)(x+1)=0(x−1)(x+1)=0

x-1=0 or x+1=0

∴ x=1 and x=-1

Since \alphaα and \betaβ are the zeroes we have that

∴ \alpha=1α=1 and \beta=-1β=−1 are the zeroes.

Now we find a quadratic polynomial whose zeroes are \frac{2\alpha}{\beta}

β

2α

and \frac{2\beta}{\alpha}

α

2β

:

For a quadratic equation from the zeroes is given by

x^2-(sum of the zeroes)x+product of the zeroes=0x

2

−(sumofthezeroes)x+productofthezeroes=0

Since the zeroes are \frac{2\alpha}{\beta}

β

2α

and \frac{2\beta}{\alpha}

α

2β

Sum of the zeroes=\frac{2\alpha}{\beta}+\frac{2\beta}{\alpha}Sumofthezeroes=

β

2α

+

α

2β

=\frac{2\alpha^2+2\beta^2}{\alpha \beta}=

αβ

2α

2

+2β

2

∴ Sum of the zeroes=\frac{2(\alpha^2+\beta^2)}{\alpha \beta}Sumofthezeroes=

αβ

2(α

2

+β

2

)

Put \alpha=1α=1 and \beta=-1β=−1 we get

Sum of the roots=\frac{2(1^2+(-1)^2)}{1(-1)}Sumoftheroots=

1(−1)

2(1

2

+(−1)

2

)

=\frac{2(2)}{-1}=

−1

2(2)

=-4=−4

∴ Sum of the zeroes=-4

Product of the zeroes=\frac{2\alpha}{\beta}\times \frac{2\beta}{\alpha}Productofthezeroes=

β

2α

×

α

2β

=4=4

∴ Product of the zeroes=4

Now we can write a quadratic equation as

x^2-(-4)x=0x

2

−(−4)x=0

x^2+4x+4=0x

2

+4x+4=0

∴ the quadratic polynomial for the zeroes \frac{2\alpha}{\beta}

β

2α

and \frac{2\beta}{\alpha}

α

2β

where \alpha=1α=1 and \beta=-1β=−1 is x^2+4x+4=0x

2

+4x+4=0 .