Question

if alpha and beta are the zeros of the quadratic polynomial 5 y square - 7y + 1 then find the value of one by Alpha square plus one by beta square​

Answer 1

Answer:

Step-by-step explanation:

5y^2-7y+1=0

alpha+beta=7/5--(1)

alpha*beta=1/5--(2)

find (1/alpha^2)+(1/beta^2)

beta^2+alpha^2/(alpha^2*beta^2)

squaring (1)

alpha^2+beta^2+2alphabeta=49/25

alpha^2+beta^2=49/25-2*(1/5)

=49/25-2/5

=49-10/25

=39/25

find

(39/25)/(1/25)=39

Answer 2

Given: 5y²-7y+1=0

find:1/alpha²+1/beta²

Solution: 5y²-7y+1=0

Comparing the above equation with

ax²+bx+c=0......... (Standard form of quadratic equation)

therefore,

a=5, b=-7 and c=1

alpha +beta=-b/a=-(-7) /5=7/5........(1)

alpha×beta=c/a=1/5 ........... (2)

1/alpha²+1/beta².........(To find)

therefore, beta²+alpha/alpha²×beta²

=(alpha+beta) ²-2×alpha×beta/alpha²×beta²

therefore, by substituting the values from (1)&(2)

=49/25-2/5

______________

(alpha×beta) ²

=49-10/25

_________

1/25

=39/25

________

1/25

=39/25×25

=39

Therefore,

1/alpha²+1/beta²=39