Question

find the quadratic polynomial whose sum and product of zeroes are√2+1 and 1/√2+1​

Answer 1

Step-by-step explanation:

Quadratic polynomial = x^2 -(√2 +1)x +(√2-1)

GIVEN:

SUM OF ZEROS = √2 +1

PRODUCT OF ZEROS = 1/(√2+1)

TO FIND:

Quadratic polynomial with the help of sum of zeros and product of zeros.

SOLUTION:

Firstly rationalise the product of zero:

\begin{gathered} = \frac{1}{ \sqrt{2} + 1} \\ = \frac{1( \sqrt{2} - 1)}{ (\sqrt{2} + 1)( \sqrt{2} - 1)} \\ = \frac{ \sqrt{2} - 1 }{2 - 1} \\ = \sqrt{2 } - 1\end{gathered}

=

2

+1

1

=

(

2

+1)(

2

−1)

1(

2

−1)

=

2−1

2

−1

=

2

−1

here

\begin{gathered} \implies \alpha + \beta \: = \sqrt{2} + 1 \\ \implies \: \alpha \: \times \beta \: = \: \sqrt{2} - 1 \\ \end{gathered}

⟹α+β=

2

+1

⟹α×β=

2

−1

Standard form of quadratic equations when sum of zeros and product of zeros are given.

x {}^{2} - ( \alpha \: + \beta)x \: + \alpha \betax

2

−(α+β)x+αβ

Putting the values:

x {}^{2} - ( \sqrt{2} + 1)x + ( \sqrt{2} - 1)x

2

−(

2

+1)x+(

2

−1)

NOTE:

Important formulas:

\begin{gathered} \implies \: \alpha \: + \beta \: = \frac{( - b)}{a} \\ \implies \: \alpha \beta \: = \frac{c}{a} \end{gathered}

⟹α+β=

a

(−b)

⟹αβ=

a

c

Where a= coefficient of x^2. b= coefficient of x.

c = constant term