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I'm really stuck on this.

Solve the inequality 4t^2 <= 9t - 2 + 13t - 5.  Write your answer in interval notation.

Apr 24, 2023

#1
+2667
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First, move everything to the left-hand side: $$4t^2 - 22t + 7 \leq 0$$

Now, let's assume that we want to solve $$4t^2 - 22t + 7 = 0$$. Using the quadratic formula, this would be $$t = {22 \pm \sqrt{22^2-4 \times 7 \times 4} \over 2 \times 4} ={ 22 \pm 2\sqrt{93} \over 8} = {11 \pm \sqrt{93} \over 4}$$

But, we haven't solved the problem yet! We need to find the values that satisfy the inequality. Note that all the points in between the interval of our zeros ($${11 - \sqrt {93} \over 4}$$ and $${11 +\sqrt {93} \over 4}$$) will either be positive of negative.

So, let's plug in $$t = 1$$ because it's in the interval. This gives us $$4(1)^2 -22 (1) + 7 = -11$$, which is less than 0. This means that the value of quadratic must be less than 0 within this interval.

In interval notation, this would be $$\color{brown}\boxed{{11 - \sqrt {93} \over 4}, {11 + \sqrt{93} \over 4}}$$

Note: There's probably a better, easier way of solving this that I can't think of.

Apr 24, 2023