Question

A quadratic polynomial has one of its zeros 1 + √5 and it satisfies p(1) = 2. Find the quadratic polynomial.

Answer 1

^_^ Quadratic Polynomial •_^

Final Answer :
${x}^{2} - (2 + \frac{7}{ \sqrt{5} } )x + ( \frac{7}{ \sqrt{5} } + 3)$

Steps:
1) Let the other zero of required polynomial p(x) be 'a'. (other than 1+ √5) .
Then,
$p(x) = (x - a)(x - (1 + \sqrt{5} ))$

We have taken leading co-efficjent as 1 , but it will not affect our result.

2) We have,
$p(1) = 2 \\ = > (1 - a)(1 - (1 + \sqrt{5} )) = 2 \\ = > (1 - a) ( - \sqrt{5} ) = 2 \\ = > a = 1 + \frac{2}{ \sqrt{5} }$

3) Therefore, Required
$p(x) = (x - (1 + \frac{2}{ \sqrt{5} } ))(x - 1 - \sqrt{5} ) \\ = > p(x) = {x}^{2} - (1 + \sqrt{5} + \frac{2}{ \sqrt{5} } + 1) + ( \frac{2}{ \sqrt{5} } + 1)(1 + \sqrt{5} ) \\ = > p(x) = {x}^{2} - (2 + \frac{7}{ \sqrt{5} } )x + ( \frac{7}{ \sqrt{5} } + 3)$


Hence, Our Required Polynomial is p(x) .

Answer 2

Step-by-step explanation:

hope it will b helpfull for u....