Question

find the polynomial whose zero are 3+√5/5,3-√5/5​

Answer 1

Given

• $\sf{Zeros \: of \: polynomial \: f(x)= \cfrac{3 + \sqrt{5} }{5}, \cfrac{3 - \sqrt{5}} {5}}$

To Find

• Polynomial

Solution

Let

$\alpha = \cfrac{3 + \sqrt{5} }{5}$

$\beta = \cfrac{3 - \sqrt{5} }{5}$

$\sf{Polynomial \: f(x)= x²-(sum \: of \: zeros)x+product \: of \: zeros}$

$\sf{Polynomial \: f(x)= x² - ( \alpha + \beta )x+ \alpha \beta }$

$\sf{Polynomial \: f(x)= x² - \left(\cfrac{3 + \sqrt{5} }{5} + \cfrac{3 - \sqrt{5} }{5} \right)x+ \cfrac{3 + \sqrt{5} }{5} \times \cfrac{3 - \sqrt{5} }{5}}$

$\sf{Polynomial \: f(x)= x² - \left(\cfrac{3 + \sqrt{5} + 3 - \sqrt{5} }{5} \right)x+ \cfrac{(3 + \sqrt{5})(3 - \sqrt{5}) }{25} }$

$\sf{Polynomial \: f(x)= x² - \left(\cfrac{3 + \cancel{\sqrt{5} }+ 3 - \cancel{\sqrt{5}} }{5} \right)x+ \cfrac{ {3}^{2} - (\sqrt{5}) ^{2} }{25} }$

$\sf{Polynomial \: f(x)= x² - \cfrac{6 }{5} x+ \cfrac{ 9 - 5 }{25} }$

$\sf{Polynomial \: f(x)= x² - \cfrac{6 }{5} x+ \cfrac{ 4 }{25} }$

$\sf{Polynomial \: f(x)= 25 \Bigg \{x² - \cfrac{6 }{5} x+ \cfrac{ 4 }{25} } \Bigg \}$

$\underbrace{\underline{\boxed{\pink{\sf{Polynomial \: f(x)= 25 x² - 30x + 4 } }}}}$