Question

find a quadratic polynomial, such that the product of 0 is 7 and one zero is 3 +root 2, ​

Answer 1

Answer:

x^2-6x+7

Step-by-step explanation:

Given that there's a quadratic Polynomial.

Also, given that,

Product of its zereos is 7.

And, one of the zeroes is (3+√2)

To find the quadratic polymonial.

Let, the other zero be x.

Therefore, we will get,

=> (3+√2)x = 7

=> x = 7/(3+√2)

Now, we have,

=> Sum of roots = (3+√2) + 7/(3+√2)

=> Sum of roots = {(3+√2)^2+7}/(3+√2)

=> Sum of roots = (9+2+6√2+7)/(3+√2)

=> Sum of roots = (18+6√2)/(3+√2)

=> Sum of roots = 6(3+√2)/(3+√2)

=> Sum of roots = 6

Now, we know that,

A quadratic polymonial having sum and product of roots m and n respectively is given by :

• x^2 -mx +n

Here, we have,

• m = 6
• n = 7

Therefore, we will get,

= x^2 -6x + 7

Hence, the required polymonial is x^2-6x+7.

Answer 2

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Let α and β are the zeroes of a quadratic polynomial.

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Given:-

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$one \:zero(α)= 3+√2$

$Sum\: of\: its\: zeroes\: (α+ β)= 6$

$(α+ β)= 6$

$3+√2 + β= 6$

$β = 6- (3+√2)= 6 -3 -√2= 3-√2$

$β= 3 -√2$

Product of zeroes (α. β) = (3+√2) (3-√2)

$α. β =( 3)² - (√2)²= 9 - 2= 7$

$(a+b) (a-b)= a² - b²$

$α. β = 7$

Required Polynomial=

k [x²-(Sum of zeroes)x +( Product of zeroes)]

= k[ x² -(α+ β)x +(α. β],

where k is a non zero real number.

$= x² -(6)x + (7) [ here k = 1]$

$= x² - 6x + 7$

Hence, a quadratic polynomial is:-

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$\implies\boxed{\bf x² - 6x +7. }$

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