Question

Find the possible number of integral values of \(x\) satisfying \(2x^2 + 11x - 138<0\)

Hint:

After finding the roots of a quadratic inequality, sketch a simple parabola. If it opens up (\(a>0\)), the function is negative between the roots. If it opens down (\(a<0\)), the function is positive between the roots. This visual check prevents errors.

Answer 1

Step 1: Understanding the Concept:
To solve a quadratic inequality of the form \(ax^2 + bx + c<0\), we first find the roots of the corresponding equation \(ax^2 + bx + c = 0\). These roots are the points where the parabola crosses the x-axis. The inequality will be satisfied for all \(x\) values between the roots if the parabola opens upwards (\(a>0\)) or outside the roots if it opens downwards (\(a<0\)).
Step 2: Detailed Explanation:
We need to solve the inequality \(2x^2 + 11x - 138<0\).
First, find the roots of the equation \(2x^2 + 11x - 138 = 0\). We use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a=2\), \(b=11\), \(c=-138\). \[ x = \frac{-11 \pm \sqrt{11^2 - 4(2)(-138)}}{2(2)} \] \[ x = \frac{-11 \pm \sqrt{121 + 8(138)}}{4} \] \[ x = \frac{-11 \pm \sqrt{121 + 1104}}{4} \] \[ x = \frac{-11 \pm \sqrt{1225}}{4} \] The square root of 1225 is 35 (since \(30^2=900\), \(40^2=1600\), and it ends in 5). \[ x = \frac{-11 \pm 35}{4} \] This gives us two roots: \[ x_1 = \frac{-11 - 35}{4} = \frac{-46}{4} = -11.5 \] \[ x_2 = \frac{-11 + 35}{4} = \frac{24}{4} = 6 \] The coefficient of the \(x^2\) term is 2, which is positive, so the parabola opens upwards. Therefore, the expression \(2x^2 + 11x - 138\) is negative (less than 0) between its roots. The solution to the inequality is: \[ -11.5<x<6 \] Step 3: Final Answer:
We need to find the number of integral values of \(x\) in this range. The integers greater than -11.5 are -11, -10, -9, ..., 0. The integers less than 6 are 5, 4, 3, ... So, the integers satisfying the inequality are \(-11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\). To count them, we use the formula: Number of integers = Last - First + 1. \[ \text{Number of integers} = 5 - (-11) + 1 = 5 + 11 + 1 = 17 \] There are 17 possible integral values for \(x\).