Suppose k ∈ R and the quadratic equation x2-(k-3) x+k=0 has at least one positive roots, then k lies in the set:

Question

Suppose k ∈ R and the quadratic equation x2-(k-3) x+k=0 has at least one positive roots, then k lies in the set: (A) (-∞, 0) ∪[9,16] (B) (-∞, 0) ∪[16,8) (C) (-∞, 0) ∪[9, ∞) (D) (-∞, 0) ∪(1, ∞)

Tardigrade Admin · Accepted Answer

Solution: Both the roots α, β will be non-positive if D ≥ 0, α+β ≤ 0, α β ≥ 0 ⇒ (k-3)2-4 k ≥ 0,(k-3) ≤ 0, k ≥ 0 ⇒ (k-1)(k-9) ≥ 0, k ≤ 3, k ≥ 0 ⇒ 0 ≤ k ≤ 1 Thus, quadratic equation will have at least one positive root if k< 0 or k >1 and (k ≤ 1 or k ≥ 9) ⇒ k ∈(-∞, 0) ∪[9, ∞)

Q. Suppose $k \in R$ and the quadratic equation $x^{2} - (k - 3) x + k = 0$ has at least one positive roots, then $k$ lies in the set:

$(- \infty, 0) \cup [9, 16]$

$(- \infty, 0) \cup [16, 8)$

$(- \infty, 0) \cup [9, \infty)$

$(- \infty, 0) \cup (1, \infty)$

Q. Suppose k∈R and the quadratic equation x2−(k−3)x+k=0 has at least one positive roots, then k lies in the set:

(−∞,0)∪[9,16]

(−∞,0)∪[16,8)

(−∞,0)∪[9,∞)

(−∞,0)∪(1,∞)

Solution: Both the roots α,β will be non-positive if D≥0,α+β≤0,αβ≥0 ⇒(k−3)2−4k≥0,(k−3)≤0,k≥0 ⇒(k−1)(k−9)≥0,k≤3,k≥0 ⇒0≤k≤1 Thus, quadratic equation will have at least one positive root if k<0 or k>1 and (k≤1 or k≥9) ⇒k∈(−∞,0)∪[9,∞)

Q. Suppose $k \in R$ and the quadratic equation $x^{2} - (k - 3) x + k = 0$ has at least one positive roots, then $k$ lies in the set:

$(- \infty, 0) \cup [9, 16]$

$(- \infty, 0) \cup [16, 8)$

$(- \infty, 0) \cup [9, \infty)$

$(- \infty, 0) \cup (1, \infty)$