Inequalities#3

Algebra Level 3

Solve the inequality:

40 x − 21 x 2 ≥ 16 \dfrac{40x-21}{x^2} \geq 16

x ∈ ( − ∞ , 0.75 ) ∪ ( 1.75 , + ∞ ) x \in (-\infty, 0.75) \cup (1.75, +\infty) x ∈ ( 0.75 , 1.75 ) x \in (0.75, 1.75) x ∈ ( − ∞ , 0.75 ] ∪ [ 1.75 , + ∞ ) x \in (-\infty, 0.75] \cup [1.75, +\infty) x ∈ [ 0.75 , 1.75 ] x \in [0.75, 1.75]

Munem Shahriar by Munem Shahriar , 18, Bangladesh

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1 solution

Chew-Seong Cheong
Jul 28, 2017

40 x − 21 x 2 ≥ 16 40 x − 21 ≥ 16 x 2 16 x 2 − 40 x + 21 ≤ 0 ( 4 x − 3 ) ( 4 x − 7 ) ≤ 0 ⟹ x ∈ [ 0.75 , 1.75 ] \begin{aligned} \frac {40x-21}{x^2} & \ge 16 \\ 40x - 21 & \ge 16x^2 \\ 16x^2 - 40x + 21 & \le 0 \\ (4x-3)(4x-7) & \le 0 \\ \implies x & \in [0.75, 1.75] \end{aligned}

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Sir, are you a staff of brilliant.org? I have noticed that when you solved this problem, you edited the options.

Munem Shahriar - 3 years, 10 months ago

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I am a moderator, helping to edit problem. It is best that you learn up LaTex.

Chew-Seong Cheong - 3 years, 10 months ago

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Thanks, I don't know it before.

Munem Shahriar - 3 years, 10 months ago

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