Question

find the zeroes of x^2 - root of 3 - x + root of 3 and verify the relationships between the two zeres and their coefficents

Answer 1

➵ p(x) = x² - √3 - x + √3

➵ p(x) = x² - x

➵ p(x) = x(x - 1)

➵ 0 = x - 1

➵ 1, 0 = x

______________________________

➵ Sum : 1 + 0 = 1

➵ Product : 1 × 0 = 0

______________________________

➵ a + ß = - Coefficient of x/Coefficient of x²

➵ aß = Constant Term/Coefficient of x²

______________________________

➵ a + ß = - (- 1)/1

➵ a + ß = 1

➵ aß = 0/1

➵ aß = 0

______________________________

Hence the relation is verified.

Answer 2

Solution :-

$\implies$ p(x) = x² - √3 - x + √3

$\implies$p(x) = x² - x

$\implies$ p(x) = x (x - 1)

Now,

$\implies$x² - √3 - x + √3 = 0

$\implies$x (x - 1) = 0

$\implies$x - 1 = 0

x = 1

x = 0

So, 1 & 0 are the zeroes of the polynomial x² - √3 - x + √3.

Verification :-

$\implies$Sum of zeroes = $\dfrac{-b}{a}$

$\implies$0 + 1 = $\dfrac{-(-1)}{1}$

$\implies$1 = 1

Also, Product of zeroes = $\dfrac{c}{a}$

$\implies$0 × 1 = $\dfrac{0}{1}$

$\implies$0 = 0

Hence, relationship between the zeroes and coefficients is verified..!!