Question

find the quadratic polynomial whose zeroes are 2 + √3 and 2 - √3​

Answer 1

we have α =2+√3 and β=2-√3

to find :- quadratic polynomial whose zeros are α and β.

let the f(x) be k( x² -(α+β)x + αβ)

now α + β = 2+√3 +2-√3

= 4

and, αβ =( 2+√3)(2-√3)

=4 -3

=1 hence th required f(x) is

k(x²-4x +1)

here k is some constant

Answer 2

Let the zeroes of the quadratic polynomial be α and β. Then, we have

α = 2 + √3

β = 2 - √3

A quadratic polynomial is of the form of ax² + bx + c (a, b and c are real numbers, a ≠ 0)

For a quadratic polynomial, the sum of zeroes =

$\sf \frac{-(coefficient\ of\ x)}{coefficient\ of\ x^2}$

Or, we can say that

α + β = - b/a

So, put the values of α and β here.

⇒ 2 + √3 + 2 - √3 = - b/a

⇒ 4 = -b/a

So, on comparing, we can say that if a = 1, b = -4.

Now, product of zeroes =

$\sf \frac{constant\ term}{coefficient\ of\ x^2}$

⇒ or, αβ = c/a

⇒ (2 + √3)(2 - √3) = c/a

⇒ 4 - 3 = c/a [since, (a + b)(a - b) = a² - b²]

⇒ 1 = c/a

So, if a = 1, c = 1

Now, put the values of a, b and c in the standard form of equation, ax² + bx + c

⇒ (1)x² + (-4)x + 1

⇒ x² - 4x + 1  

which is, the required polynomial.

$\rule{200}2$

Alternative Method

A quadratic equation is also given as :-

k[x² - (sum of roots)x + (product of roots)]

where, k is a constant term. For k = 1, the polynomial would be

x² - (sum of roots)x + (product of roots)

Here, sum of roots = 4 and product of roots = 1. So the polynomial would be :-

x² - 4x + 1

For k = 2, the polynomial would become

2(x² - 4x + 1)

= 2x² - 8x + 2

which is another polynomial having the same roots.