How do you solve the inequality $9x^2+16>=24x$ and write your answer in interval notation?

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Answer 1 · 2017-05-25T12:12:46

Solution: $x in RR$ , In Interval notation: $(-oo,oo)$

$9x^2+16>= 24x or 9x^2+16- 24x >=0 or 9x^2- 24x+16 >=0$ or

$(3x-4)^2 >=0 or (3x-4)(3x-4) >=0$ . Critical point is $x = 4/3$

At $x =4/3 ; (3x-4)(3x-4) =0$ .On either side of $x=4/3 ; (3x-4)(3x-4) >0 :. x in RR$ , In Interval notation: $(-oo,oo)$

Solution: $x in RR$ , In Interval notation: $(-oo,oo)$
graph{9x^2-24x+16 [-10, 10, -5, 5]} [Ans]

How do you solve the inequality 9x^2+16>=24x9x^2+16>=24x and write your answer in interval notation?