Question

If alpha and beta are the zeroes of the polynomial 3x^2 + 5x -2 then find the values of ( alpha-beta)^2​

Answer 1

Answer:

$\sf{The \ value \ of \ (\alpha-\beta)^{2} \ is \ \dfrac{49}{9}.}$

Given:

$\sf{\leadsto{If \ \alpha \ and \ \beta \ are \ the \ zeroes}} \\ \sf{of \ the \ polynomial \ 3x^{2}+5x-2}$

To find:

$\sf{The \ value \ of \ (\alpha-\beta)^{2}.}$

Solution:

$\sf{The \ given \ polynomial \ is \ 3x^{2}+5x-2} \\ \sf{Here, \ a=3, \ b=5, \ c=-2} \\ \\ \sf{\alpha \ and \ \beta \ are \ zeroes.} \\ \\ \sf{\alpha+\beta=\dfrac{-b}{a}} \\ \\ \sf{\therefore{\alpha+\beta=\dfrac{-5}{3}...(1)}} \\ \\ \sf{\alpha\beta=\dfrac{c}{a}} \\ \\ \sf{\therefore{\alpha\beta=\dfrac{-2}{3}...(2)}} \\ \\ \boxed{\sf{(a-b)^{2}=(a+b)^{2}-4ab}} \\ \\ \sf{\therefore{(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta}} \\ \\ \sf{from \ (1) \ and \ (2), \ we \ get} \\ \\ \sf{(\alpha-\beta)^{2}=(\dfrac{-5}{3})^{2}-4\times(\dfrac{-2}{3})} \\ \\ \sf{\therefore{(\alpha-\beta)^{2}=\dfrac{25}{9}+\dfrac{8}{3}}} \\ \\ \sf{\therefore{(\alpha-\beta)^{2}=\dfrac{25+24}{9}}} \\ \\ \sf{\therefore{(\alpha-\beta)^{2}=\dfrac{49}{9}}} \\ \\ \purple{\tt{\therefore{The \ value \ of \ (\alpha-\beta)^{2} \ is \ \dfrac{49}{9}.}}}$