I've been trying to factorize this quadratic equation to find it's roots. (I know how to use discriminant but this has to be solved using factorization)
I asssumed the middle term will split in terms of √5 and then I'm stuck at another quadratic. Can't figure out it's roots without using discriminant. Any help?
On
It is all a matter of presentation I suppose. Here is one approach:
The roots solve $x^2-\frac{\sqrt 5 }{2} x-1=0$. The vertex is at $\sqrt 5/4,$ and since the roots are equidistant (call it a distance $t$) from the vertex and multiply to $-1$, we have $(\sqrt 5/4-t)(\sqrt 5/4+t)=-1\implies 5/16-t^2=-1\implies t=\sqrt {21}/4,$
implying the roots are $\frac{\sqrt 5\pm\sqrt{21}}{4}$.
You could use the quadratic complement, that is, divide the equation by the coefficient of $x^2$, then add and subtract the square of half of the new coefficient of $x$.
So let us firstly divide the given equation by $2$, yielding $$x^2-\frac{\sqrt 5}{2}x-1=0.$$ Then we add and subtract $5/16$ to the left member: $$x^2-\frac{\sqrt 5}{2}x+\frac{5}{16}-\frac{5}{16}-1=0.$$ Tidying up gives $$\Big(x-\frac{\sqrt 5}{4}\Big)^2=\frac{21}{16}.$$ Can you finish from here?