Question

√3x^2 +√2x-2√3=0 solve by quadratic formula method of 10th standard​

Answer 1

Answer:

The solution for the equation $\sqrt{3}x^{2} +\sqrt{2}x-2\sqrt{3}=0$ is $x=1.064$ and $x=-1.878$

Step-by-step explanation:

The question is to solve the equation using the quadratic formula. The given equation is$\sqrt{3}x^{2} +\sqrt{2}x-2\sqrt{3}=0$. This equation is in the form$ax^{2} +bx+c=0$. The quadratic formula for this type of equation is as follows,    

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

Here, $a=\sqrt{3} , b=\sqrt{2}$ and $c=-2\sqrt{3}$

Put these values in the quadratic formula. Then,

$x=\frac{-\sqrt{2} }{2\sqrt{3} }$ ± $\frac{\sqrt{\sqrt{2} ^{2}-4.\sqrt{3}.(-2\sqrt{3}) } }{2\sqrt{3} }$

If the value of $b^{2}-4ac$ is zero, then the quadratic equation will have one real root, if it is greater than zero then the quadratic will have 2 real roots, and if it is less than zero the quadratic equation will have no real roots.

$x=\frac{-\sqrt{2} }{2\sqrt{3} }$ ± $\frac{\sqrt{2+24} }{2\sqrt{3} }$

Here $b^{2}-4ac$ is greater than zero, then there will have 2 real roots.

That is,

$x=-0.407$ ± $1.471$

$x=-0.407+1.471=1.064$ and $x=-0.407-1.471=-1.878$

This is the solution to the question.