Solution set of |x2-5 x+7|+|x2-5 x-14|=21 is

Question

Solution set of |x2-5 x+7|+|x2-5 x-14|=21 is (A) [-2,7] (B) (-∞,-2] ∪[2, ∞) (C) [7, ∞) (D) (-∞,-2]

Tardigrade Admin · Accepted Answer

|x2-5 x+7|+|4+5 x-x2|=21 |x|+|y|=|x+y| x y ≥ 0 (x2-5 x+7)(14+5 x-x2) ≥ 0 (x2-5 x+7)(x2-5 x-14) ≤ 0 x2-5 x=t (t+7)(t-14) ≤ 0 t ∈[-7,14] -7 ≤ x2-5 x ≤ 14

Q. Solution set of $∣ ∣ x^{2} - 5 x + 7 ∣ ∣ + ∣ ∣ x^{2} - 5 x - 14 ∣ ∣ = 21$ is

$[- 2, 7]$

$(- \infty, - 2] \cup [2, \infty)$

$[7, \infty)$

$(- \infty, - 2]$

Q. Solution set of ∣∣​x2−5x+7∣∣​+∣∣​x2−5x−14∣∣​=21 is

[−2,7]

(−∞,−2]∪[2,∞)

[7,∞)

(−∞,−2]

∣∣​x2−5x+7∣∣​+∣∣​4+5x−x2∣∣​=21 ∣x∣+∣y∣=∣x+y∣ xy≥0 (x2−5x+7)(14+5x−x2)≥0 (x2−5x+7)(x2−5x−14)≤0 x2−5x=t (t+7)(t−14)≤0 t∈[−7,14] −7≤x2−5x≤14

Q. Solution set of $∣ ∣ x^{2} - 5 x + 7 ∣ ∣ + ∣ ∣ x^{2} - 5 x - 14 ∣ ∣ = 21$ is

$[- 2, 7]$

$(- \infty, - 2] \cup [2, \infty)$

$[7, \infty)$

$(- \infty, - 2]$