Question

If ‘a’ and ‘b’ are the zeroes of the quadratic polynomial x2-3x+2; find the value of 1/a²+1/b²

Answer 1

Answer:

1/(2α+β) and 1/(α+2β)

Step-by-step explanation:

if α & β are the zeroes of the quadratic polynomial f(x) =x² -3x-2, then how do you find a quadratic polynomial whose zeroes are 1/ (2 α + β) and 1/ (α +2 β)?

α,β are roots of x2−3x−2

⟹α+β=3;αβ=−2

(2α+β)(α+2β)=2α2+2β2+5αβ

=2(α+β)2+αβ=2(32)+(−2)=16

For the required quadratic equation, roots are12α+β,1α+2β

⟹Sum of the roots = 12α+β+1α+2β=2α+β+2β+α(2α+β)(α+2β)

=3(α+β)(2α+β)(α+2β)=3(3)16=916

⟹Product of the roots = 12α+β⋅1α+2β=116

∴Required quadratic equation = X2−916X+116=0

⟹16X2−9X+1=0

Your quadratic polynomial is f(x)=x^2–3x-2 , any quadratic equation have 2 roots ( equalor unequal or conjugate) , Since α,β are zeros of the f(x) ,so α and β are the roots of equation ,x^2- 3x -2=0 ; so sum of roots α+β = -(-3)/ 1 [ as for general equation ;ax^2+bx +c =0 ; sum of roots =-b/a , prrodct of roots = c/a ] and product of roots αβ =-2/1 ,so α+β = 3 and αβ = -2 ,

Again Equation of any Quadratic Equation is ; x^2 - ( sum of roots) x + product of the roots = 0 ;

Given 1/(2α+β) and 1/(α+2β) are zeros of quadratic polynomial, then the roots of such polynomial must be 1/(2α+β) and 1/(α+2β) , Hence Equation of the Polynomial is ; x^2 - {1/(2α+β) + 1/(α+2β)}x + {1/( 2α+β)}{1/(α+2β)} =0 , So x^2 - {1/(3+α)+ 1/(3+β)}x + 1/ { (3+α)(3+β)} =0 [ since α+β=3 ] or (3+α)(3+β) x^2 - { (3+α) +(3+β)} x +1 =0 or { 9+ 3(α+β) +αβ}x^2 - {6+(α+β)}x +1 =0 [since αβ=-2 ] or ( 9+ 3,3 -2))x^2 - (6+3)x +1=0 or 16x^2 -9x +1 =0 , Therefore the polynomial is f(x) = 16x^2 -9x +1 whose zeros are 1/(2α+β) and 1/(α+2β) : Answer