Question

koe tou bata dou eska answer step by step

21 question​

Answer 1

Given equation

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

Solving, we get,

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

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

$\implies 3x(x+\sqrt{5})-\sqrt{5} (x+\sqrt{5})$

$\implies (3x-\sqrt{5})(x+\sqrt{5})$

Now,

$3x-\sqrt{5}= 0$

$\implies x = \dfrac{\sqrt{5} }{3}$

$x+\sqrt{5}=0\\ \implies x = -\sqrt{5}$

$\boxed{\boxed{\bold{ Therefore, \ the \ zeroes \ of \ the \ polynomial \ are \ \frac{\sqrt{5} }{3} \ and \ -\sqrt{5}}}}}}}$

                                                       

Answer 2

Answer:

upper answer is correct brother