Question

6. Find the quadratic equation whose zeroes are 3 + √2 and 3-√2.​

Answer 1

Given zeros of polynomial are 3±
2
​

Let α,β be zeros of polynomial
then
α+β=3+
2
​
+3−
2
​
=6 ________ (1)
& αβ(3+
2
​
)(3−
2
​
)
αβ=9−3
2
​
+3
2
​
−2=7 _________ (2)
then
quadratic equation is
x
2
−(α+β)x+αβ=0
From (1) & (2)
[x
2
−6x+7=
Hope it helps

Answer 2

Given : Zeroes of a quadratic equation are ( 3 + √2) and (3 - √2)

To find : Quadratic equation

Solution :

Every quadratic equation can be expressed in terms of its zeroes as ::

x² - ( sum of zeroes ) x + Product

In order to find the quadratic equation, we have to find the sum and product of its zeroes.

Sum of zeroes

Sum = ( 3 + √2) + (3 - √2)

Sum = 3 + √2 + 3 - √2

Sum = 3 + 3 + √2 - √2

Sum = 6

Product of zeroes

Product = ( 3 + √2 ) ( 3 - √2 )

Product = (3)² - (√2)²

Product = 9 - 2

Product = 7

[ Here we used the identity (A+B)(A-B) = A² - B² ]

Quadratic equation

Quadratic equation = x² - (sum)x + Product

Quadratic equation = x² - 6x + 7

Hence the required answer is x² - 6x + 7

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