Question

if alpha and beta are the zeros of the quadratic polynomial x2-x-6 find a quadratic polynomial whose zeros are [alpha +beta] and 1/alpha + 1/beta

Answer 1

x2-x-6=0
x²-3x+2x-6=0
x(x-3)+2(x-3)=0
(x-3)(x+2)=0
=@=3,bita=-2
alpha+bita=3+(-2)=1
1/alpha+1/bita=1/3-1/2=2-3/6=
eq.is x²+(a1+a2)+a1a2=0
x²-(1-1/6)x+1(-1/6)=0
x²+5/6x-1/6=0
6x²-5x-1=0

Answer 2

Answer:

Quadratic polynomial is $k(x^2)-\frac{-5}{6}x+(\frac{-1}{6})=k(6x^2+5x-1)$

Step-by-step explanation:

Given Quadratic Polynomial ,

x² - x - 6

we first find zeroes of the fiven polynomial,

x² - x - 6 = 0

x² - 3x + 2x - 6 = 0

x ( x - 3 ) + 2 ( x - 3 ) = 0

( x - 3 ) ( x + 2 ) = 0

⇒ x = 3 , -2

⇒ α = 3 & β = -2

So, Zeroes of New polynomial,

α + β = 3 + ( -2 ) = 1

1/α + 1/β = 1/3 + 1/(-2) = 1/(-6)

let say zeroes of new polynomial,

x = 1 and y = 1/(-6)

Sum of zeroes of new polynomial = 1 +  1/(-6) = 5/(-6)

Product of zeroes = 1 × 1/(-6) = 1/(-6)

Quadratic polynomial is $k(x^2)-\frac{-5}{6}x+(\frac{-1}{6})=k(6x^2+5x-1)$