Question

if alpha and beta are the zeros of 2x^2+3x+5 find alpha^2/beta +beta^2/alpha​

Answer 1

Given :

• ${\small{\tt{\alpha \ and \ \beta \ are \ the \ zeroes \ of \ polynomial \ 2x^2 + 3x + 5}}}$

To Find :

• ${\small{\tt{ Find \ : \ \ \ \ \dfrac{\alpha ^2}{\beta} + \dfrac{\beta ^2}{\alpha}}}}$

Solution:

${\small{\tt{We \ know \ that ,}}}$

${\small{\tt{\alpha + \beta = - \dfrac{coefficient \ of \ x}{coefficient \ of \ x^2}}}}$

${\rightarrow{\small{\tt\bold{ \pink{\alpha + \beta = - \dfrac{3}{2}}}}}}$

${\small{\tt{\alpha\beta = \dfrac{constant \ term}{coefficient \ of \ x^2}}}}$

${\rightarrow{\small{\tt\bold{\pink{\alpha\beta= \dfrac{5}{2}}}}}}$

${\small{\tt{Now,}}}$

${\small{\tt{\dfrac{\alpha}{\beta} + \dfrac{ \beta}{\alpha}}}}$

${\small{\tt{ \: = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}}}}$

${\small{\tt{ = \dfrac{ (\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}}}}$

${\small{\tt{= \dfrac{\bigg(- \dfrac{3}{2}\bigg)^2 - 2 \bigg(\dfrac{5}{2} \bigg)}{\dfrac{5}{2}}}}}$

${\small{\tt{ = \dfrac{\dfrac{9}{4} - 5}{\dfrac{5}{2}}}}}$

${\small{\tt{ = \dfrac{\dfrac{9-20}{4}}{\dfrac{5}{2}}}}}$

${\small{\tt{ = \dfrac{-\dfrac{11}{4}}{\dfrac{5}{2}}}}}$

${\small{\tt{ = -\dfrac{11}{4} × \dfrac{2}{5}}}}$

${\small{\tt = { \underline{ \boxed{ -\dfrac{11}{10}}}}}}$

${\large{ \red{\therefore \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{-11}{10}}}}$

Answer 2

Answer:

The polynomial can be factored into: (x+1)(2x-5), so we have α = -1 and β = 5/2

(i) The polynomial with zeros α2 and β2 can be factored as (x - α2)(x - β2). This can be multiplied out to get

x2 - (α2 + β2)x + α2β2 or x2 - (1 + 25/4)x + 25/4 or x2 - (29/4)x + 25/4

Multiply all the terms by 4 to get 4x2 - 29x + 25

See if you can now do part (ii).

Step-by-step explanation:

Hope it will help you..