6.246. The quadratic equation $a x^{2}+b x+c=0$ has two roots. Form a new quadratic equation, one of whose roots is one unit less than the larger root, and the other is one unit more than the smaller root of the given equation.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0392.jpg?height=123&width=1121&top_left_y=1497&top_left_x=142)
The smaller root of the quadratic equation \(a x^{2}+b x+c=0\).

Then \(y_{1}=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}-1, y_{2}=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}+1\) are the roots of the quadratic equation \(y^{2}+p x+q=0\).

By Vieta's theorem, we have

\[
\begin{aligned}
& \left\{\begin{array}{l}
y_{1}+y_{2}=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}-1+\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}+1=-\frac{b}{a}, \\
y_{1} \cdot y_{2}=\left(\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}-1\right)\left(\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}+1\right)=\frac{c-a+\sqrt{b^{2}-4 a c}}{a}
\end{array}\right. \\
& \left\{\begin{array}{l}
\frac{b}{a}=p \\
q=\frac{c-a+\sqrt{b^{2}-4 a c}}{a}, \text { from which we have } y^{2}+\frac{b}{a} y+\frac{c-a+\sqrt{b^{2}-4 a c}}{a}=0 \Leftrightarrow
\end{array}\right. \\
& a y^{2}+b y+\left(c-a+\sqrt{b^{2}-4 a c}\right)=0
\end{aligned}
\]

Answer: \(a y^{2}+b y+\left(c-a+\sqrt{b^{2}-4 a c}\right)=0\).