Question

if alpha, beta are the roots of the quadratic equation ax^2+bx+c=0 find a quadratic equation whose roots are alpha^2+beta^2 and aloha^-2+beta^-2​

Answer 1

Answer:

a²c²x² - ( b^2 - 2ac )( a^2 + c^2 )x + ( b^2 - 2ac )^2 = 0

Step-by-step explanation:

Eq. is ax^2 + bx + c = 0    ⇒ x^2 - (-b/a)x + (c/a) = 0.

Quadratic equation written in the form of x^2 - Sx + P = 0 represent S as sum of roots and P as the product of their roots.

So, here if α and β are roots.

S = α + β = - b / a  

P = αβ = c / a

Therefore,

α^2 + β^2 = (α+β)² - 2αβ

    = ( - b / a )^2 - 2( c / a )

    = b^2 / a^2 - 2c / a

    = ( b^2 - 2ac ) / a^2

1 / α^2 + 1 / β^2

= ( α^2 + β^2 ) / ( αβ )^2

= {(b^2 - 2ac)/a^2}/{(c/a)^2}

= ( b^2 - 2ac ) / c^2

Here, we are asked for the equation with roots ( b^2 - 2ac ) / a^2 and ( b^2 - 2ac ) / c^2.

 So, if that equation is x^2 - Kx + L = 0. So, K is the sum of roots and L is the product of roots.

This means :

K = ( b^2 - 2ac ) / a^2 + ( b^2 - 2ac ) / c^2

K = ( b^2 - 2ac )( a^2 + c^2 ) / ( ac )^2

 And,

L = ( b^2 - 2ac )/a^2 * ( b^2 - 2ac )/c^2

L = ( b^2 - 2ac )^2 / ( ac )^2

Hence the required equation is :

 x^2 - x( b^2 - 2ac )( a^2 + c^2 )/(ac)^2 + ( b^2 - 2ac )^2/(ac)^2 = 0

⇒ a²c²x² - ( b^2 - 2ac )( a^2 + c^2 )x + ( b^2 - 2ac )^2 = 0

Hence the required equation is a²c²x² - ( b^2 - 2ac )( a^2 + c^2 )x + ( b^2 - 2ac )^2 = 0.

Answer 2

Given:-

$\alpha\ and\ \beta$ are the roots of the quadratic equation $ax^2+bx+c$.

To find:-

The quadratic equation whose roots are $\alpha ^2\ and\ \beta ^2\ AND\ \frac{1}{\alpha ^2}\ and\ \frac{1}{\beta ^2}$.

$(\alpha ^-^2=\frac{1}{\alpha ^2}\ and\ \beta ^-^2=\frac{1}{\beta ^2})$

Solution:-

The quadratic equation ax² + bx + c = 0 has 2 roots, i.e. α and β.

The quadratic equation can also be written as :

x² - (Sum of the zeros)x + (Product of the zeros) = 0

• α + β(sum) = -b/a

• αβ(product) = c/a

∴ x² - (-b/a)x + (c/a) = 0

1. Similarly, the given zeros of the required quadratic equation are :

α² and β².

Sum of the zeros = α² + β² :

= (α + β)² - 2αβ

= (-b/a)² - 2 c/a

= (b² - 2 ac)/a²

Product of the zeros :

α²β²

= (αβ)²

= (c/a)²

= c²/a²

∴Hence, the required polynomial is k{x² - [(b² - 2ac)/a]x + c²/a²} = 0

2. Again, the given zeros are 1/α² and 1/β².

Sum of the zeros :

1/α² + 1/β²

= (α² + β²)/α²β²

= [(α + β)² - 2αβ]/(αβ)²

= [(-b/a)² - 2 c/a]/(c/a)²

= (b²/a² - 2 c/a)/(c²/a²)

= b² - 2 ac/a² * a²/c²

= b² - 2 ac/c²

Product of the zeros :

1/α²*1/β²

= 1/(αβ)²

= 1/(c/a)²

= a²/c²

∴Hence, the required polynomial is k{x² - [(b² - 2 ac)/c²]x + a²/c²} = 0