Question

Solve the following quadratic equation:
$x^2-3\sqrt{5}x+10=0$

Answer 1

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

We know that general form of quadratic equation is $a{x}^2+bx+c=0$

On comparing, $a=1,b=-3\sqrt{5}$ and $c=10$

Now,

$b=-3\sqrt{5}x=-2\sqrt{5}x-\sqrt{5}x$
$ac=x^2\times 10=10x^2=-2\sqrt{5}x\times -\sqrt{5}x$

So, we can write $-3\sqrt{5}x$ as $-2\sqrt{5}x-\sqrt{5}x$

Now, $x^2-3\sqrt{5}x+10=0$
$\Rightarrow x^2-2\sqrt{5}x-\sqrt{5}x+2\sqrt{5}\times \sqrt{5}=0$
$\Rightarrow x(x-2\sqrt{5})-\sqrt{5}(x-2\sqrt{5})=0$
$\Rightarrow (x-2\sqrt{5})(x-\sqrt{5})=0$
$\Rightarrow (x-2\sqrt{5})=0\text{and}(x-\sqrt{5})=0$
$\Rightarrow x=2\sqrt{5}\text{and}x=\sqrt{5}$

Hence, the roots of the given equation are $2\sqrt{5}$ and $\sqrt{5}$.