The number of real roots of the equation x ∣ x − 2 ∣ + 3 ∣ x − 3 ∣ + 1 = 0 is:
Question: The number of real roots of the equation x ∣ x − 2 ∣ + 3 ∣ x − 3 ∣ + 1 = 0 is: 📷 Question Image: Short Solution (Text): Critical points for the absolute values are x = 2 x=2 and x = 3 x=3 . Solve piecewise in the three intervals. 1) Region x < 2 x<2 : ∣ x − 2 ∣ = 2 − x , ∣ x − 3 ∣ = 3 − x |x-2|=2-x,\; |x-3|=3-x . The equation becomes x ( 2 − x ) + 3 ( 3 − x ) + 1 = 0 ⇒ − x 2 − x + 10 = 0 ⇒ x 2 + x − 10 = 0. x(2-x) + 3(3-x) + 1 = 0 \\ \Rightarrow -x^2 - x + 10 = 0 \\ \Rightarrow x^2 + x - 10 = 0. Discriminant Δ = 1 + 40 = 41 \Delta = 1 + 40 = 41 . Roots: x = − 1 ± 41 2 . x = \frac{-1 \pm \sqrt{41}}{2}. Numeric values: − 1 + 41 2 ≈ 2.7016 \dfrac{-1+\sqrt{41}}{2}\approx 2.7016 and − 1 − 41 2 ≈ − 3.7016 \dfrac{-1-\sqrt{41}}{2}\approx -3.7016 . Only x ≈ − 3.7016 x\approx -3.7016 lies in this region ( x < 2 x<2 ) → one valid root here. 2) Region 2 ≤ x < 3 2 \le x < 3 : ∣ x − 2 ∣ = x − 2 , ∣ x − 3 ∣ = 3 − x |x-2|=x-2,\; |x-3|=3-x . The e...