Question

If $\alpha and \beta$ are the zeros of the quadratic polynomial $f(t)= t^{2} -4t+3$,find the value of $\alpha^{4} \beta^{3}+ \alpha^{3} \beta^{4}$

Answer 1

SOLUTION :

Given : α and β are the zeroes of the quadratic polynomial f(t)= t² - 4t + 3

On comparing with at² + bt + c,

a = 1 , b= -4 , c= 3

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = -(-4)/1 = 4

α + β = 4 …………………….(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = 3/1 = 3

αβ =  3 ……………………(2)

Now,

α⁴β³ +α³β⁴ = α³β³(α + β) = (αβ)³(α + β)

α⁴β³ +α³β⁴ = (αβ)³(α + β)

By Substituting the value from eq 1 & 2

= 3³ × 4 = 27 × 4  =   108

α⁴β³ +α³β⁴ = 108

Hence, the value of α⁴β³ +α³β⁴ is 108.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer:

Step-by-step explanation:

Alpha be = d

Beta be =e

d^4×e^3+d^3×e^4

(de)^3(d+e). (Taking d^3×e^3 common)

Quadratic equation is of form

ax^2+bx+c

And Sum of roots = -b/a

= -(-4)/1=4

Product of roots = c/a

= 3/1=3

(de)^3(d+e)=3^3×4= 27× 4= 108