Question

if alpha and beta are the zeros of polynomia 4square-5x-1 find the value of alpha square beta +beta square alpha​

Answer 1

Answer:

4x²-5x+-1=0

α,β are the zeroes

⇒α+β= -b/a

⇒α+β= -(-5)/4=5/4

also, αβ=c/a= -1/4

Now, α²β+β²α= αβ(α+β)

        -1/4(5/4)= -5/16

✌

Answer 2

Answer:

$\sf{The \ value \ of \ \alpha^{2}\beta+\beta^{2}\alpha \ is \ \dfrac{-5}{16}.}$

Given:

$\sf{\alpha \ and \ \beta \ are \ the \ zeroes \ of \ the} \\ \\ \sf{polynomial \ 4x^{2}-5x-1}$

To find:

$\sf{The \ value \ of \ \alpha^{2}\beta+\beta^{2}\alpha.}$

Solution:

$\sf{p(x)=4x^{2}-5x-1} \\ \\ \sf{Here, \ a=4, \ b=-5 \ and \ c=-1} \\ \\ \boxed{\sf{Sum \ of \ zeroes=\dfrac{-b}{a}}} \\ \\ \sf{\therefore{\alpha+\beta=\dfrac{5}{4}...(1)}} \\ \\ \boxed{\sf{Product \ of \ zeroes=\dfrac{c}{a}}} \\ \\ \sf{\therefore{\alpha\beta=\dfrac{-1}{4}...(2)}}$

$\sf{Now,} \\ \\ \sf{\longmapsto{\alpha^{2}\beta+\beta^{2}\alpha}} \\ \\ \sf{\longmapsto{\alpha\beta(\alpha+\beta)}} \\ \\ \sf{\longmapsto{\dfrac{-1}{4}(\dfrac{5}{4})}} \\ \\ \sf{...from \ (1) \ and \ (2)} \\ \\ \sf{\longmapsto{\dfrac{-5}{16}}}$

$\purple{\tt{\therefore{The \ value \ of \ \alpha^{2}\beta+\beta^{2}\alpha \ is \ \dfrac{-5}{16}.}}}$