find a quadratic polynomial ,the sum and the product of zeroes are √2 and 1/√2
Answers
Step-by-step explanation:
ANSWER:
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{lgathered}= \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{lgathered}
=
2
+1
1
=
(
2
+1)(
2
−1)
1(
2
−1)
=
2−1
2
−1
=
2
−1
here
\begin{lgathered}\implies \alpha + \beta \: = \sqrt{2} + 1 \\ \implies \: \alpha \: \times \beta \: = \: \sqrt{2} - 1 \\\end{lgathered}
⟹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{lgathered}\implies \: \alpha \: + \beta \: = \frac{( - b)}{a} \\ \implies \: \alpha \beta \: = \frac{c}{a}\end{lgathered}
⟹α+β=
a
(−b)
⟹αβ=
a
c
Where a= coefficient of x^2. b= coefficient of x.
c = constant term
Answer:
2x^2-2√2x+√2
Step-by-step explanation:
General Form Of Quadratic Polynomial
=x^2-(sum of zeroes)x+(product of zeroes)
=x^2-(√2)x+(1/√2)
=x^2-(√2)x+(√2/2)
=2(x^2-√2x+√2/2)
=2x^2-2√2x+√2
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