Show that the roots of $x^2 − 2α x +β = 0$ are $x = α \pm \sqrt {α^2 −β}$

Question

I have been asked to show that the roots of $$x^2-2αx+β$$ are $$x=α±\sqrt{α^2- β}$$

I do not have any idea how to complete this question, I believe factorising is involved but I don't know how to apply it if so.

I have got the steps here but I am not able to complete the square. please can someone show me? this is what I have.

Any help in terms of finding a solution and showing methodology would be very much appreciated.

Edited let us solve this equation

$x^2+3*x-4=0$

let us calculate discriminant $d=b^2-4*ac=9+16=25$

$x_1=(-b+\sqrt(d))/2=(-3+5)/2=1$

$x_2=(-3-5)/2)=-4$

Answer 1

Hint : Vieta's theorem allows a quick proof. The sum of the roots must be $2\alpha$ and the product $\beta$

Answer 2

$d=4*\alpha^2-4*\beta$

which is the same as $4*(\alpha^2-\beta) $ square root from this will be $2*\sqrt{\alpha^2-\beta)}$

can you continue?

we will have ($2*\alpha+2*\sqrt{\alpha^2-\beta)}) /2$

and
($2*\alpha-2*\sqrt{a^2-\beta)}) /2$

factorize $2$ out of bracket

Answer 3

Consider $f(x) = x^2-2\alpha x+\beta = 0$

$f(x) = (x-\alpha)^2+\beta-\alpha^2 =0$

Then $x= \alpha\pm\sqrt{\alpha^2-\beta}$ as required.

Answer 4

These are simply the so-called reduced formulae for quadratic equations $\;ax^2+2b'x+c=0\,$: one sets $$\Delta'=b'^2-ac\qquad(\textit{reduced discriminant}).$$ If $\Delta'> 0$, the (real) root are $$x_1, x_2=\frac{-b'\pm\sqrt{\Delta'}}{a}.$$