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Chapter 4: Problem 20

The solution of the inequation, \(15 x^{2}-31 x+14<0\) is given by (1) \(\mathrm{x} \in\left(\frac{7}{5}, \infty\right)\) (2) \(\frac{2}{3}

Short Answer

Expert verified
Question: Determine the interval(s) for which the inequality \(15x^2 - 31x + 14 < 0\) is true. Answer: The inequality is true for the interval \(\frac{2}{3} < x < \frac{7}{5}\).

Step by step solution

01

Factor the quadratic expression

The quadratic expression \(15x^2 - 31x + 14\) can be factored as \((3x-2)(5x-7)\).
02

Find the roots

To find the roots, we set each factor equal to zero and solve for \(x\): \begin{align*} 3x-2 & =0\\ x&=\frac{2}{3}, \end{align*} and \begin{align*} 5x-7&=0\\ x&=\frac{7}{5}. \end{align*}
03

Test intervals

We divide the number line into three intervals based on the roots \(x = \frac{2}{3}\) and \(x=\frac{7}{5}\): (1) \(x < \frac{2}{3}\) (2) \(\frac{2}{3} < x < \frac{7}{5}\) (3) \(x > \frac{7}{5}\) We will test a number from each interval in the factored inequality \((3x-2)(5x-7) < 0\): (1) Let \(x = 0\) (in interval 1); \((3(0)-2)(5(0)-7) = (2)(7) > 0\), which does not satisfy the inequality. (2) Let \(x = 1\) (in interval 2); \((3(1)-2)(5(1)-7) = (1)(-2) < 0\), which satisfies the inequality. (3) Let \(x = 2\) (in interval 3); \((3(2)-2)(5(2)-7) = (4)(3) > 0\), which does not satisfy the inequality.
04

State the solution

Based on our tests, we find that the solution to the inequality \((15x^2 - 31x + 14) < 0\) is \(\frac{2}{3} < x < \frac{7}{5}\), which corresponds to option (2).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Quadratics
When solving quadratic inequalities, one essential step is factoring the quadratic expression. In our example, the quadratic expression is \(15x^2 - 31x + 14\). Factoring helps break down the expression into simpler components, making it easier to solve. The goal is to express the quadratic as a product of two linear factors.

For our expression, this results in \((3x-2)(5x-7)\). This means that the quadratic can be rewritten as this multiplication of binomials. Factoring quadratics often requires practice to recognize patterns or might involve using techniques like the quadratic formula or completing the square if straightforward factoring isn鈥檛 apparent.

When correctly factored, as in this case, it simplifies solving the inequality to checking these factors individually.
Roots of Quadratic Equation
Once we have a factored quadratic expression, the next step is to find its roots. The roots are the values of \(x\) which make the expression equal to zero. They represent the points where the graph of the quadratic equation intersects the x-axis.

For the expression \((3x-2)(5x-7)\), setting each factor to zero individually will help us find the roots:
  • Solving \(3x-2=0\) gives \(x=\frac{2}{3}\)
  • Solving \(5x-7=0\) gives \(x=\frac{7}{5}\)
These roots, \(x=\frac{2}{3}\) and \(x=\frac{7}{5}\), divide the number line into different regions, important for solving inequalities. Understanding these roots helps determine where the quadratic expression changes the sign on the number line, which is crucial for interval testing.
Interval Testing
After determining the roots, interval testing is used to solve the inequality. This process involves selecting test points from each interval divided by the roots to determine where the original inequality holds true.

Given the roots \(x = \frac{2}{3}\) and \(x = \frac{7}{5}\), we divide the number line into three intervals:
  • Interval 1: \(x < \frac{2}{3}\)
  • Interval 2: \(\frac{2}{3} < x < \frac{7}{5}\)
  • Interval 3: \(x > \frac{7}{5}\)
We pick a number from each interval (for example, 0 for Interval 1, 1 for Interval 2, and 2 for Interval 3), substitute into the factored inequality \((3x-2)(5x-7) < 0\), and check the sign of the result.

Through testing, we establish that the inequality holds only in Interval 2, \(\frac{2}{3} < x < \frac{7}{5}\). This means the quadratic expression is negative in this range, satisfying the inequality.

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Most popular questions from this chapter

Find the roots of the equation \(\ell^{2}\left(\mathrm{~m}^{2}-\mathrm{n}^{2}\right) \mathrm{x}^{2}+\mathrm{m}^{2}\left(\mathrm{n}^{2}-\ell^{2}\right) \mathrm{x}+\mathrm{n}^{2}\left(\ell^{2}-\mathrm{m}^{2}\right)=0\) (1) \(1, \frac{\mathrm{n}^{2}\left(\ell^{2}-\mathrm{m}^{2}\right)}{\ell^{2}\left(\mathrm{~m}^{2}-\mathrm{n}^{2}\right)}\) (2) \(1, \frac{-\mathrm{m}^{2}\left(\ell^{2}-\mathrm{n}^{2}\right)}{\ell^{2}\left(\mathrm{~m}^{2}-\mathrm{n}^{2}\right)}\) (3) \(1, \frac{\mathrm{n}^{2}\left(\ell^{2}+\mathrm{m}^{2}\right)}{\ell^{2}\left(\mathrm{~m}^{2}-\mathrm{n}^{2}\right)}\) (4) \(1, \frac{-m^{2}\left(\ell^{2}+n^{2}\right)}{\ell^{2}\left(m^{2}-n^{2}\right)}\)

In a forest, a certain number of apes equal to the square of one-eighth of the total number of their group are playing and having great fun. The rest of them are twelve in number and are on an adjoining hill. The echo of their shrieks from the hills frightens them. They come and join the apes in the forest and play with enthusiasm. What is the total number of apes in the forest? (1) 16 (2) 48 (3) 16 or 48 (4) 64

The roots of the equation \(x^{2}-p x+q=0\) are consecutive integers. Find the discriminant of the equation. (1) 1 (2) 2 (3) 3 (4) 4

If \(y^{2}+6 y-3 m=0\) and \(y^{2}-3 y+m=0\) have a common root, then find the possible values of \(m\). (1) \(0,-\frac{27}{16}\) (2) \(0,-\frac{81}{16}\) (3) \(0, \frac{81}{16}\) (4) \(0, \frac{27}{16}\)

If \(a-b, b-c\) are the roots of \(a x^{2}+b x+c=0\), then find the value of \(\frac{(a-b)(b-c)}{c-a}\). (1) \(\frac{\mathrm{b}}{\mathrm{c}}\) (2) \(\frac{\mathrm{c}}{\mathrm{b}}\) (3) \(\frac{a b}{c}\) (4) \(\frac{\mathrm{bc}}{\mathrm{a}}\)
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