Roots of the equation $ (1): ax^2+bx+c=0$ are $x_{1}$ and $x_{2}$. They are both real.
Without solving first equation, make up new quadratic equation such that one of the its roots is $x_{1} + 1$ and second one is $x_{2}-1$. Note that $x_{2}>x_{1}$.
I started to solve this with Vieta's theorem but I couldn't continue. Deep explanation would be grateful!
Also I don't know what does not solve mean, I can't use a quadratic formula or something more specific?
Best Answer
$f(x)=ax^2+bx+c=a(x-x_1)(x-x_2)$ is given to you. Consider \begin{align*} g(x)&=a(x-(x_1+1))(x-(x_2-1))\\ &=a(x-x_1-1)(x-x_2+1)\\ &=a(\color{red}{(x-x_1)}-1)(\color{blue}{(x-x_2)}+1)\\ &=a(x-x_1)(x-x_2)+a(x-x_1)-a(x-x_2)-a\\ &=f(x)+a\underbrace{(x_2-x_1)}_{\text{given }>0}-a. \end{align*} Now $x_2-x_2=\sqrt{(x_1+x_2)^2-4x_1x_2}=\sqrt{\left(\frac{b^2}{a^2}\right)-\frac{4c}{a}}=\frac{\sqrt{b^2-4ac}}{a}$ Thus $$g(x)=ax^2+bx+c+\sqrt{b^2-4ac}-a$$