Question

Find the quadratic polynomial whose
zeroes are 3+√5 and 3-√5
​

Answer 1

Answer:

Answer

Let α=3+

5

and β=3

5

Sum of zeros =α+β=3+

5

+3−

5

=6

Product of zeros =αβ=(3+

5

)(3−

5

)=9−5=4

Also, Sum of roots =

a

−b

=−

1

6

Product of roots =

a

c

=

1

4

⟹a=1,b=−6 and c=4

Hence, the quadratic polynomial is x

2

−6x+4

Step-by-step explanation:

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Answer 2

Zeroes of Quadrant potential $\sf\mathbb\color{red} {⟹(3 + \sqrt{5}) } \: and \: (3 - \sqrt{5} )$

Let $\sf\mathbb\color{red} { \alpha = (3 + \sqrt{5} )} \: and \: \beta = (3 - \sqrt{5} )$

We know that,

Quadratic polynomial

$\sf\mathbb\color{blue} { = {x}^{2} - (sum \: of \: zeroes)x \: + (product \: of \: zeroes \: ) }$

Sum of zeros :

$\sf\mathbb\color{red} {⟹ \alpha + \beta }$

$\sf\mathbb\color{red} {⟹(3 + \sqrt{5}) } + (3 - \sqrt{5} )$

$\sf\mathbb\color{red} {⟹3 + \sqrt{5} + 3 - \sqrt{5} }$

$\sf\mathbb\color{red} {⟹6}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\underline{\bf { \alpha + \beta = 6 }}}$

Product of zeroes:

$\sf\mathbb\color{green} { \alpha \beta }$

$\sf\mathbb\color{green} {(3 + \sqrt{3})(3 - \sqrt{5} )}$

$\sf\mathbb\color{green} {⟹ {3}^{2} - {( \sqrt{5}) }^{2} }$

$\sf\mathbb\color{green} {⟹}9 - 5$

$\sf\mathbb\color{green} {⟹}4$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\underline{\bf { \alpha \beta = 4 }}}$

• Therefore, the Quadratic polynomial

$\sf\mathbb\color{purple} {⟹ {x}^{2} - ( \alpha + \beta )x + \alpha \beta }$

$\sf\mathbb\color{purple} {⟹ {x}^{2} - 6x + 4}$