Question

If the sum of zeros is -3/2√5 and the product of zeros is -1/2. then find the zeros of the polynomial​

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ zeroes \ of \ the \ polynomial \ are}$

$\sf{\frac{1}{\sqrt5} \ and \ \frac{-\sqrt5}{2}.}$

$\sf\orangr{Given:}$

$\sf{\implies{Sum \ of \ zeroes=\frac{-3}{2\sqrt5}}}$

$\sf{\implies{Product \ of \ zeroes=\frac{-1}{2}}}$

$\sf\pink{To \ find:}$

$\sf{Zeroes \ of \ the \ polynomial.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ zeroes \ of \ the \ polynomial}$

$\sf{be \ \alpha \ and \ \beta.}$

$\sf{Sum \ of \ zeroes=\frac{-3}{2\sqrt5}}$

$\sf{\therefore{\alpha+\beta=\frac{-3}{2\sqrt5}...(1)}}$

$\sf{Product \ of \ zeroes=\frac{-1}{2}}$

$\sf{\therefore{\alpha\beta=\frac{-1}{2}...(2)}}$

$\sf{According \ to \ the \ identity.}$

$\sf{(a-b)^{2}=(a+b)^{2}-4ab}$

$\sf{\therefore{(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta}}$

$\sf{....from \ (1) \ and \ (2)}$

$\sf{\therefore{(\alpha-\beta)^{2}=(\frac{-3}{2\sqrt5})^{2}-4\times(\frac{-1}{2})}}$

$\sf{\therefore{(\alpha-\beta)^{2}=\frac{9}{20}+2}}$

$\sf{\therefore{(\alpha-\beta)^{2}=\frac{9+40}{20}}}$

$\sf{\therefore{(\alpha-\beta)^{2}=\frac{49}{20}}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{\alpha-\beta=\frac{7}{2\sqrt5}...(3)}$

$\sf{Add \ equations \ (1) \ and \ (3)}$

$\sf{\alpha+\beta=\frac{-3}{2\sqrt5}}$

$\sf{+}$

$\sf{\alpha-\beta=\frac{7}{2\sqrt5}}$

_______________________

$\sf{2\alpha=\frac{-3+7}{2\sqrt5}}$

$\boxed{\sf{\therefore{\alpha=\frac{1}{\sqrt5}}}}$

$\sf{Substitute \ \alpha=\frac{1}{\sqrt5} \ in \ equation (2)}$

$\sf{\therefore{\frac{1}{\sqrt5}\times\beta=\frac{-1}{2}}}$

$\boxed{\sf{\therefore{\beta=\frac{-\sqrt5}{2}}}}$

$\sf\purple{\tt{\therefore{The \ zeroes \ of \ the \ polynomial \ are}}}$

$\sf\purple{\tt{\frac{1}{\sqrt5} \ and \ \frac{-\sqrt5}{2}.}}$