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:

### Related Formula X^2 = |X|^2$X^2 = |X|^2$ Quadratic factorization: t^2 - 5t + 6 = (t-2)(t-3)$t^2 - 5t + 6 = (t-2)(t-3)$ ### Core Logic Rewrite the equation taking |x - 1| = t$|x - 1| = t$, where t geq 0$t \geq 0$. Since (x-1)^2 = |x-1|^2$(x-1)^2 = |x-1|^2$, the equation becomes: t^2 - 5t + 6 = 0$$t^2 - 5t + 6 = 0$$ ### Step 1: Solve for modulus (t - 2)(t - 3) = 0 Rightarrow t = 2, 3$$(t - 2)(t - 3) = 0 \Rightarrow t = 2, 3$$ Since both roots are positive, both provide valid solutions for the modulus. |x - 1| = 2 quad textand quad |x - 1| = 3$|x - 1| = 2 \quad \text{and} \quad |x - 1| = 3$ ### Step 2: Unpack x values From |x - 1| = 2$|x - 1| = 2$: x - 1 = 2 Rightarrow x = 3$$x - 1 = 2 \Rightarrow x = 3$$ x - 1 = -2 Rightarrow x = -1$$x - 1 = -2 \Rightarrow x = -1$$ From |x - 1| = 3$|x - 1| = 3$: x - 1 = 3 Rightarrow x = 4$$x - 1 = 3 \Rightarrow x = 4$$ x - 1 = -3 Rightarrow x = -2$$x - 1 = -3 \Rightarrow x = -2$$ The roots are 3, -1, 4, -2$3, -1, 4, -2$. ### Step 3: Sum of Roots textSum = 3 + (-1) + 4 + (-2) = 4$$\text{Sum} = 3 + (-1) + 4 + (-2) = 4$$ ### Pattern Recognition In symmetric modulus equations f(|x-a|) = 0$f(|x-a|) = 0$, every valid root t$t$ generates twin solutions (a+t)$(a+t)$ and (a-t)$(a-t)$. The sum of each pair is perfectly 2a$2a$. If there are n$n$ distinct valid positive roots for t$t$, the sum of all x$x$-roots is exactly n times 2a$n \times 2a$. Here, n=2, a=1 Rightarrow 2 times 2(1) = 4$n=2, a=1 \Rightarrow 2 \times 2(1) = 4$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations