The sum of all the roots of the equation (x-1)²-5|x-1|+6=0$(x-1)^{2}-5|x-1|+6=0$ , is:

Core Logic

For the denominator x² - 8x + 32$x^2 - 8x + 32$, D = 64 - 128 < 0$D = 64 - 128 < 0$ and a = 1 > 0$a = 1 > 0$. Thus, x² - 8x + 32 > 0 x in R$x^2 - 8x + 32 > 0 \forall x \in \mathbb{R}$.

Step 1: Constraint on Numerator

Since the denominator is always positive, the numerator must be strictly negative for all x in R$x \in \mathbb{R}$.

This requires a < 0$a < 0$ and D < 0$D < 0$.

Step 2: Conclusion

Since a$a$ must be strictly less than 0$0$, there are no positive integral values of a$a$ that satisfy the condition. Hence, S$S$ is an empty set. Number of elements is 0.

Chapter Mix

Class 11 Maths: Quadratic Equations