What is the solution set for #4x^2 - 5x < 6#?

Bring the inequality to standard form: #f(x) = 4x^2 - 5x - 6 < 0# First, solve #f(x) = 4x^2 - 5x - 6 = 0# (1) to get the 2 real roots. I use the new Transforming Method. (Google, Yahoo) Transformed equation #f'(x) = x^2 - 5x + 24# (2). Roots have opposite signs. Factor pairs of 24 -> ...(-2, 12)(-3, 8). This sum is 5 = -b. Then, the 2 real roots of (2) are: -3 and 8. Back to original equation (1), the 2 real roots are: #-3/4# and #8/4 = 2.# Find the solution set of the inequality. Since a > 0, the parabola opens upward. Between the 2 real roots #(-3/4)# and (2), a part of the parabola is below the x-axis, meaning f(x) < 0. Answer by open interval:# (-3/4, 2)#

To find the solution set for the inequality \(4x^2 - 5x < 6\), we first rearrange the inequality to set it equal to zero. Then, we solve the quadratic equation using methods like factoring, completing the square, or using the quadratic formula to find the critical points. Finally, we determine the intervals where the inequality holds true. The quadratic equation \(4x^2 - 5x - 6 = 0\) can be solved using factoring, completing the square, or the quadratic formula. Using the quadratic formula: \[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\] where \(a = 4\), \(b = -5\), and \(c = -6\). \[x = \frac{{5 \pm \sqrt{{(-5)^2 - 4 \cdot 4 \cdot (-6)}}}}{{2 \cdot 4}}\] \[x = \frac{{5 \pm \sqrt{{25 + 96}}}}{{8}}\] \[x = \frac{{5 \pm \sqrt{{121}}}}{{8}}\] \[x = \frac{{5 \pm 11}}{{8}}\] So, the solutions are \(x = \frac{{5 + 11}}{{8}}\) and \(x = \frac{{5 - 11}}{{8}}\): \[x_1 = \frac{{16}}{{8}} = 2\] \[x_2 = \frac{{-6}}{{8}} = -\frac{{3}}{{4}}\] Now, we have the critical points \(x = 2\) and \(x = -\frac{{3}}{{4}}\). We test the intervals \(x < -\frac{{3}}{{4}}\), \(-\frac{{3}}{{4}} < x < 2\), and \(x > 2\) by plugging in test values into the original inequality \(4x^2 - 5x < 6\). We find that the inequality holds true for \(x < -\frac{{3}}{{4}}\) and \(x > 2\), but not for \(-\frac{{3}}{{4}} < x < 2\). So, the solution set for \(4x^2 - 5x < 6\) is \(x < -\frac{{3}}{{4}}\) or \(x > 2\).