The equation |1-x|+x^2=5 has

$$\begin{aligned} & |1-x|+x^2=5 \\ & \Rightarrow|x-1|+x^2=5 \end{aligned}$$ First case: If $x < 1,|x-1|$ is negative. $$\begin{aligned} & \therefore \quad-(\mathrm{x}-1)+\mathrm{x}^2=5 \\ & \Rightarrow-\mathrm{x}+1+\mathrm{x}^2=5 \\ & \Rightarrow \mathrm{x}^2-\mathrm{x}+1=5 \\ & \Rightarrow \mathrm{x}^2-\mathrm{x}-4=0 \\ & \text{ Roots are } \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-4)}}{2(1)} \end{aligned}$$=\frac{1 \pm \sqrt{1+16}}{2}=\frac{1 \pm \sqrt{17}}{2}$Since,$x < 1$, root cannot be$\frac{1+\sqrt{17}}{2}$. So, the root is$\frac{1-\sqrt{17}}{2}$, which is irrational. Second case: If$x > 1,|x-1|=x-1$$\begin{array}{ll} \therefore & |\mathrm{x}-1|+\mathrm{x}^2=5 \\ \Rightarrow & \mathrm{x}-1+\mathrm{x}^2=5 \\ \Rightarrow & \mathrm{x}^2+\mathrm{x}-6=0 \\ \Rightarrow & \mathrm{x}^2+3 \mathrm{x}-2 \mathrm{x}-6=0 \\ \Rightarrow & \mathrm{x}(\mathrm{x}+3)-2(\mathrm{x}+3)=0 \\ \Rightarrow & \mathrm{x}=2,-3 \end{array}$$ Since $x > 1$, root cannot be -3 . So, root is 2 which is rational. $\therefore \quad$ given expression has one irrational root and one rational root.$