Question

PZ SOLVE IT FOR ME PLZ​

Answer 1

Step-by-step explanation:

Given :-

If α and β are the zeroes of the polynomial ax²+bx+c .

To find :-

Find the other polynomial whose zeroes are

α²/ β and β²/α ?

Solution :-

Given quadratic polynomial P(x) = ax²+bx+c

Given zeroes = α and β

We know that

Sum of the zeroes = -b/a

=> α + β = -b/a -----------(1)

Product of the zeroes = c/a

=> α β = c/a -------------(2)

We know that

The quadratic polynomial whose zeores are α and β is K[x²-(α+β)x+αβ]

Now sum of the required zeroes

(α²/ β) +( β²/α)

=>(α³+ β³)/α β

=>[ (α +β)³-3(α+β)(α β)]/α β

Since (a+b)³ = a³+b³+3ab(a+b)

From (1) & (2)

=> [(-b/a)³-3(-b/a)(c/a)]/(c/a)

==> [(-b/a)³+3(bc/a²)]/(c/a)

=> [(-b³/a³)+(3bc/a²)]/(c/a)

=>[ (-b³+3abc)/a²]/(c/a)

=>[ (-b³+3abc)/a²]×(a/c)

=> (-b³+3abc)(c)/a

=> (-b³c+3abc²)/a

and

Product of the required zeroes

=> (α²/ β) ×( β²/α)

=> (α²β²) ×( βα)

=> αβ

From (2)

=> c/a

The quadratic polynomial whose zeores are α²/ β and β²/α

=> K [x²-({-b³c+3abc²}/a)x+(c/a)]

=> K[x²+(-b³c+3abc²)/a}x+(c/a)]

=> K[{ax²+(3abc²-b³c)x+c}/a]

If K = a then the required Polynomial is

ax²+(3abc²-b³c)x+c

Answer:-

The quadratic polynomial whose zeores are α²/ β and β²/α is ax²+(3abc²-b³c)x+c

Used formulae:-

• The standard quadratic polynomial is ax²+bx+c
• Sum of the zeroes = -b/a
• Product of the zeroes = c/a
• The quadratic polynomial whose zeores are α and β is K[x²-(α+β)x+αβ]
• (a+b)³ = a³+b³+3ab(a+b)