4. Given the quadratic function $y=x^{2}-x+a$ whose graph intersects the $x$-axis at two distinct points, the sum of the distances from these points to the origin does not exceed 5. Then the range of values for $a$ is $\qquad$ .
4. $-6 \leqslant a0$, i.e., $a<\frac{1}{4}$, and $\left|x_{1}\right|+$ $\left|x_{2}\right| \leqslant 5$. Therefore,
$$
\begin{array}{l}
\left(\left|x_{1}\right|+\left|x_{2}\right|\right)^{2}=x_{1}^{2}+x_{2}^{2}+2\left|x_{1} x_{2}\right| \\
=\left(x_{1}+x_{2}\right)^{2}+2\left|x_{1} x_{2}\right|-2 x_{1} x_{2} \\
=1+2|a|-2 a \leqslant 25 \text {. } \\
\text { And }\left\{\begin{array}{l}
a<\frac{1}{4}, \\
1+2|a|-2 a \leqslant 25
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array} { l } 
{ 0 \leqslant a < \frac { 1 } { 4 } , } \\
{ 1 \leqslant 2 5 }
\end{array} \text { or } \left\{\begin{array}{l}
a<0, \\
-4 a \leqslant 24
\end{array}\right.\right. \\
\Leftrightarrow 0 \leqslant a<\frac{1}{4} \text { or }-6 \leqslant a<0 \Leftrightarrow-6 \leqslant a<\frac{1}{4} \text {. } \\
\end{array}
$$