$x^2-10x+10=0$, If two roots of this equation are $\alpha$ and $\beta(\alpha>\beta)$,find the value of$(\alpha-\beta)^2$.

Can someone help me with this question, and maybe add some explanation? I've looked at the other answers on this website about Quadratic questions, but I just don't quite understand it. I've also looked on Wikipedia, but it doesn't help either. Thanks!

In general if $\alpha$ and $\beta$ are the roots of equation $ax^2+bx+c=0$, then,

Sum of the roots, $\alpha+\beta=-\dfrac{\text{coefficient of }x}{\text{coefficient of }x^2}=-\dfrac{b}{a}$, and

Multiplication of roots, $\alpha\beta=\dfrac{\text{constant term}}{\text{coefficient of }x^2}=\dfrac{c}{a}$.

Similarly as $\alpha$ and $\beta$ are the roots of equation $x^2-10x+10=0$

so, $\alpha+\beta=10$ and, $\alpha\beta=10$.

now, \begin{align*} (\alpha-\beta)^2&=(\alpha+\beta)^2-4\alpha\beta\\ &=10^2-4\times10\\ &=100-40\\ &=60 \end{align*}

• I've edited the ans. please check @Anna – k.Vijay Apr 25 '17 at 12:06
• Thanks for helping me! Especially with the explanation! :) – Anna Apr 25 '17 at 13:46

Hint: $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$

With the usual formula for quadratic equations you get that the solutions are

$$5 \pm \sqrt{15}.$$

Quadratic equations may have $2$,$1$ or none real roots.It depends of the value of discriminant(Δ). If Δ$>0$ it has $2$ roots, if Δ$=0$ it has $1$ and if it is smaller than $0$ it does not have any. The formula for Δ $= b^2-4ac$, where $a$ is the coefficient of $x^2$, $b$ the coefficient of $x$ and $c$ the constant. Then i hope, you know the formula for the roots. Then you let $a$ equal the biggest of these roots and $b$ the other one and you find the value of $(a-b)^2$