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Question: $\frac{3x^2}{5-4x} < 4$ $-x-4$...

3x25−4x<4\frac{3x^2}{5-4x} < 4 −x−4-x-4

Answer

(−8−2313,−8+2313)∪(54,∞)\left(\frac{-8 - 2\sqrt{31}}{3}, \frac{-8 + 2\sqrt{31}}{3}\right) \cup \left(\frac{5}{4}, \infty\right)

Explanation

Solution

The given inequality is 3x25−4x<4\frac{3x^2}{5-4x} < 4. To solve this, we move all terms to one side:

3x25−4x−4<0\frac{3x^2}{5-4x} - 4 < 0

Combine the terms on the left side into a single fraction:

3x2−4(5−4x)5−4x<0\frac{3x^2 - 4(5-4x)}{5-4x} < 0

3x2−20+16x5−4x<0\frac{3x^2 - 20 + 16x}{5-4x} < 0

3x2+16x−205−4x<0\frac{3x^2 + 16x - 20}{5-4x} < 0

Now, we find the roots of the numerator and the denominator. For the numerator, 3x2+16x−20=03x^2 + 16x - 20 = 0. Using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}:

x=−16±162−4(3)(−20)2(3)x = \frac{-16 \pm \sqrt{16^2 - 4(3)(-20)}}{2(3)}

x=−16±256+2406x = \frac{-16 \pm \sqrt{256 + 240}}{6}

x=−16±4966x = \frac{-16 \pm \sqrt{496}}{6}

x=−16±16×316x = \frac{-16 \pm \sqrt{16 \times 31}}{6}

x=−16±4316x = \frac{-16 \pm 4\sqrt{31}}{6}

x=−8±2313x = \frac{-8 \pm 2\sqrt{31}}{3}

Let the roots of the numerator be r1=−8−2313r_1 = \frac{-8 - 2\sqrt{31}}{3} and r2=−8+2313r_2 = \frac{-8 + 2\sqrt{31}}{3}.

For the denominator, 5−4x=05-4x = 0.

4x=54x = 5

x=54x = \frac{5}{4}

Let the root of the denominator be r3=54r_3 = \frac{5}{4}.

We need to order these critical points on the number line. 31\sqrt{31} is between 5 and 6, approximately 5.57.

r1=−8−2313≈−8−2(5.57)3=−8−11.143=−19.143≈−6.38r_1 = \frac{-8 - 2\sqrt{31}}{3} \approx \frac{-8 - 2(5.57)}{3} = \frac{-8 - 11.14}{3} = \frac{-19.14}{3} \approx -6.38

r2=−8+2313≈−8+2(5.57)3=−8+11.143=3.143≈1.05r_2 = \frac{-8 + 2\sqrt{31}}{3} \approx \frac{-8 + 2(5.57)}{3} = \frac{-8 + 11.14}{3} = \frac{3.14}{3} \approx 1.05

r3=54=1.25r_3 = \frac{5}{4} = 1.25

The ordered critical points are r1<r2<r3r_1 < r_2 < r_3.

The inequality is 3x2+16x−205−4x<0\frac{3x^2 + 16x - 20}{5-4x} < 0. Let f(x)=3(x−r1)(x−r2)−(4x−5)=3(x−r1)(x−r2)−4(x−r3)f(x) = \frac{3(x-r_1)(x-r_2)}{-(4x-5)} = \frac{3(x-r_1)(x-r_2)}{-4(x-r_3)}. This can be written as f(x)=−34(x−r1)(x−r2)(x−r3)f(x) = -\frac{3}{4} \frac{(x-r_1)(x-r_2)}{(x-r_3)}. We want f(x)<0f(x) < 0, which means −34(x−r1)(x−r2)(x−r3)<0-\frac{3}{4} \frac{(x-r_1)(x-r_2)}{(x-r_3)} < 0. Multiplying by −43-\frac{4}{3} (which is negative) reverses the inequality sign:

(x−r1)(x−r2)(x−r3)>0\frac{(x-r_1)(x-r_2)}{(x-r_3)} > 0.

We use the wavy curve method or test intervals for the expression (x−r1)(x−r2)(x−r3)\frac{(x-r_1)(x-r_2)}{(x-r_3)}. The critical points are r1,r2,r3r_1, r_2, r_3. The critical points divide the number line into four intervals: (−∞,r1)(-\infty, r_1), (r1,r2)(r_1, r_2), (r2,r3)(r_2, r_3), and (r3,∞)(r_3, \infty).

Let g(x)=(x−r1)(x−r2)(x−r3)g(x) = \frac{(x-r_1)(x-r_2)}{(x-r_3)}.

  • For x>r3x > r_3: (x−r1)>0(x-r_1) > 0, (x−r2)>0(x-r_2) > 0, (x−r3)>0(x-r_3) > 0. So g(x)=(+)(+)(+)>0g(x) = \frac{(+)(+)}{(+)} > 0.

  • For r2<x<r3r_2 < x < r_3: (x−r1)>0(x-r_1) > 0, (x−r2)>0(x-r_2) > 0, (x−r3)<0(x-r_3) < 0. So g(x)=(+)(+)(−)<0g(x) = \frac{(+)(+)}{(-)} < 0.

  • For r1<x<r2r_1 < x < r_2: (x−r1)>0(x-r_1) > 0, (x−r2)<0(x-r_2) < 0, (x−r3)<0(x-r_3) < 0. So g(x)=(+)(−)(−)>0g(x) = \frac{(+)(-)}{(-)} > 0.

  • For x<r1x < r_1: (x−r1)<0(x-r_1) < 0, (x−r2)<0(x-r_2) < 0, (x−r3)<0(x-r_3) < 0. So g(x)=(−)(−)(−)<0g(x) = \frac{(-)(-)}{(-)} < 0.

We are looking for the intervals where g(x)>0g(x) > 0. These are (r1,r2)(r_1, r_2) and (r3,∞)(r_3, \infty). Substituting the values of r1r_1, r2r_2, and r3r_3:

The solution is x∈(−8−2313,−8+2313)∪(54,∞)x \in \left(\frac{-8 - 2\sqrt{31}}{3}, \frac{-8 + 2\sqrt{31}}{3}\right) \cup \left(\frac{5}{4}, \infty\right).