Question

find the roots of following quadratic equation: 15x² -10√6x + 10

Answer 1

[15x^2-10√6x+10=0]by5
3x^2-2√6+2=0
D=b^2-4ac
  =(-2√6)^2 - 4·3·2
  =24-24
  =0
x=(-b±√D)/2a
  =[(--2√6)±√0]/2·3
  =2√6/6
  =√2/√3
Therefore roots are √2/√3 and √2/√3.

Answer 2

The roots of the quadratic equation  $15x^{2} -10\sqrt{6} x + 10=0$  are  $\frac{\sqrt{2} }{\sqrt{3} } \ and \ \frac{\sqrt{2} }{\sqrt{3} }$ .

Step-by-step explanation:

Given:

The quadratic equation $15x^{2} -10\sqrt{6} x + 10=0$.

To  Find:

The roots of the quadratic equation $15x^{2} -10\sqrt{6} x + 10=0$.

Formula Used:

The quardatic equation $ax^{2} +bx+c=0$.

The roots of quardatic equation  are  given by

$x =\frac{-b \± \sqrt{b^{2} -4ac} }{y}$    ----- equation no.01.

Solution:

As given- the quadratic  equation $15x^{2} -10\sqrt{6} x + 10=0$.

Compared with the quadratic equation $ax^{2} +bx+c=0$.

$a= 15 , b= -10\sqrt{6 } \ and \ c=10$.

Putting value of a,b and c in formula no.01.

$x =\frac{-(-10\sqrt{6} ) \± \sqrt{(-10\sqrt{6} )^{2} -4\times15\times10} }{2\times 15}$

   $=\frac{10\sqrt{6} \± \sqrt{600 - 600} }{30}$

   $=\frac{10\sqrt{6} \± 0 }{30}$

   $=\frac{10\sqrt{6} \ }{30}$

   $=\frac{\sqrt{6} \ }{3}$

   $=\frac{\sqrt{2} \times\sqrt{3} }{\sqrt{3} \times \sqrt{3} }$

   $=\frac{\sqrt{2} }{\sqrt{3} }$

$x=\frac{\sqrt{2} }{\sqrt{3} } \ and \ \frac{\sqrt{2} }{\sqrt{3} }$

Thus, the roots of quadratic equation  $15x^{2} -10\sqrt{6} x + 10=0$ are  $\frac{\sqrt{2} }{\sqrt{3} } \ and \ \frac{\sqrt{2} }{\sqrt{3} }$ .