Question

SOLVE question number 14
I want immediate and correct answer

Answer 1

hope this helps you ☺☺

Answer 2

$Given : f(x) = x^3 - 3 \sqrt{5} x^2 + 13x - 3 \sqrt{5}$

$Given : g(x) = x - \sqrt{5}$

Given that g(x) is a factor of p(x).

$x^3 - \sqrt{5} x^2 - 2 \sqrt{5} x^2 + 10x + 3x - 3 \sqrt{5} = 0$

$(x - \sqrt{5})x^2 - (x - \sqrt{5} )2 \sqrt{5} x + (x - \sqrt{5} ) * 3 = 0$

$(x - \sqrt{5} )(x^2 - 2 \sqrt{5} x + 3) = 0$

(1)

$x - \sqrt{5} = 0$

$x = \sqrt{5}$


(2)

$x^2 - 2 \sqrt{5} + 3 = 0$

Using roots of quadratic equation, we get

$a = 1, b = 2 \sqrt{5} , c = 3$


(i)

$x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}$

          $= \frac{-b + \sqrt{(2 \sqrt{5})^2 - 4 * 1 * 3 } }{2 * 1}$

          $\frac{-(-2 \sqrt{5}) + \sqrt{20 - 12} }{2}$

          $\frac{2 \sqrt{5} + \sqrt{8} }{2}$

          $\frac{2( \sqrt{2} + \sqrt{5}) }{2}$

          $\sqrt{2} + \sqrt{5}$



(ii)

$x = \frac{-b- \sqrt{b^2 - 4ac} }{2a}$

          $= \frac{-(2 \sqrt{5}) - \sqrt{20-12} }{2}$

          $= \frac{2 \sqrt{5} - \sqrt{8} }{2}$

          $= \frac{2 \sqrt{5} - 2 \sqrt{2} }{2}$

          $\frac{2( \sqrt{5} - \sqrt{2} ) }{2}$

          $\sqrt{5} - \sqrt{2}$



Therefore the zeroes of the polynomial are:

$\sqrt{5} , \sqrt{5} + \sqrt{2} , \sqrt{5} - \sqrt{2}$



Hope this helps!