if alpha & beta are zeros of polynomial kx²+5x+2.such that 1/alpha²

1.	if alpha & beta are zeros of polynomial kx²+5x+2.such that 1/alpha²+1/beta² = 17/4.find the value of k
Answer» α + ß = \(\frac{-b}a=\frac{-5}k\) αß = \(\frac ca=\frac 2k\) Now, \(\frac1{\alpha^2} + \frac1{\beta^2}\) = \(\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}=\frac{(α+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}\) = \(\cfrac{(\frac{-5}k)^2-4/k}{(\frac 2k)^2}\) = \(\cfrac{\frac{25}{k^2}-\frac 4k}{\frac4{k^2}}\) = \(\frac{25-4k}4=\frac{17}4\) ⇒ 25 - 4k = 17 ⇒ 4k = 25 - 17 = 8 ⇒ k = 2

if alpha & beta are zeros of polynomial kx²+5x+2.such that 1/alpha²+1/beta² = 17/4.find the value of k

Answer»

α + ß = \(\frac{-b}a=\frac{-5}k\)

αß = \(\frac ca=\frac 2k\)

Now, \(\frac1{\alpha^2} + \frac1{\beta^2}\) = \(\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}=\frac{(α+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}\)

= \(\cfrac{(\frac{-5}k)^2-4/k}{(\frac 2k)^2}\) = \(\cfrac{\frac{25}{k^2}-\frac 4k}{\frac4{k^2}}\)

= \(\frac{25-4k}4=\frac{17}4\)

⇒ 25 - 4k = 17

⇒ 4k = 25 - 17 = 8

⇒ k = 2