Question

a and ß are zeroes of 2x² + x -5 find a² + ß² ii) a²/ß + ß²/a.

Answer 1

Answer:

hope it helps.........

Answer 2

Answer:

$(\alpha )^2+(\beta )^2=\dfrac{21}{4}\\\\\\ \dfrac{(\alpha )^2}{\beta }+ \dfrac{(\beta )^2}{\alpha }=\dfrac{31}{20}$

Step-by-step explanation:

${Given \ p(x)=2x^2 + x -5}\\\\For \ sum\\\\\alpha+ \beta=\dfrac{-b}{a}\\\\\\\alpha +\beta=\dfrac{-1}{2} \ ...(i)\\\\\\For \ product \\\\\\\alpha \beta =\dfrac{c}{a}\\\\\\\alpha \beta =\dfrac{-5}{2}\\\\\\we \ know \ that\\\\\\Squaring \ (i)\\\\\\(\alpha +\beta)^2=(\dfrac{-1}{2})^2\\\\\\ (\alpha )^2+(\beta )^2=\dfrac{1}{4} -2\times\dfrac{-5}{2}\\\\\\(\alpha )^2+(\beta )^2=\dfrac{21}{4}$

ii.

$\dfrac{(\alpha )^2}{\beta }+ \dfrac{(\beta )^2}{\alpha } =\dfrac{(\alpha +\beta )((\alpha )^2+(\beta )^2-\alpha \beta)}{\alpha\beta }\\\\\\putting \ values \ here\\\\\\\dfrac{(\alpha )^2}{\beta }+ \dfrac{(\beta )^2}{\alpha } =(\dfrac{\frac{-1}{2})(\frac{21}{4} +\frac{5}{2})}{\frac{-5}{2} })\\\\\\ \dfrac{(\alpha )^2}{\beta }+ \dfrac{(\beta )^2}{\alpha }=\dfrac{62}{40}\\\\\\ \dfrac{(\alpha )^2}{\beta }+ \dfrac{(\beta )^2}{\alpha }=\dfrac{31}{20}$