Question

If a and ß are the zeroes of 4x2–
6X+2 find the value of a³+ß³​

Answer 1

Step-by-step explanation:

Given: α,β are zeroes of the polynomial 4x2+3x+7

To find the value of α1+β1

From the given quadratic equation,

α+β=−43 and αβ=47

Therefore, 

α1+β1=αβα+β

=47−43=−73

⇒α1+β1=−73

Answer 2

Answer:

alpha=1/2,beta=1 therefore ,

a^3+beta^3=(1/2)^3+(1)=9/8