Solving $7x^2 - 2x leqslant 5$ A Step-by-Step Guide

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In the realm of mathematics, quadratic inequalities often present a challenge to students. Understanding the steps involved in solving these inequalities is crucial for mastering algebra. This article provides a detailed, step-by-step solution to the quadratic inequality $7x^2 - 2x leqslant{leqslant} 5$, ensuring clarity and comprehension for readers of all levels. We will break down the problem, explain the underlying concepts, and guide you through the process to arrive at the correct solution. Our goal is to make this seemingly complex problem accessible and easy to understand. Let's dive in and conquer this mathematical hurdle together!

Understanding Quadratic Inequalities

Before we tackle the specific problem, it's essential to understand what quadratic inequalities are and the general approach to solving them. A quadratic inequality is an inequality that involves a quadratic expression, which is a polynomial expression of degree two. This means the highest power of the variable (in our case, x) is two. Quadratic inequalities often take the form of $ax^2 + bx + c > 0$, $ax^2 + bx + c < 0$, $ax^2 + bx + c \geqslant 0$, or $ax^2 + bx + c \leqslant 0$, where a, b, and c are constants, and a is not equal to zero.

The key to solving quadratic inequalities lies in understanding the behavior of quadratic functions. A quadratic function, when graphed, forms a parabola. The solutions to the inequality depend on whether the parabola opens upwards (if a > 0) or downwards (if a < 0) and where it intersects the x-axis. These intersection points, also known as the roots or zeros of the quadratic equation, play a crucial role in determining the intervals where the inequality holds true.

To solve a quadratic inequality, we generally follow these steps:

  1. Rearrange the inequality: Bring all terms to one side, leaving zero on the other side. This standard form allows us to easily identify the coefficients and work with the quadratic expression. For instance, if we have $7x^2 - 2x \leqslant 5$, we need to subtract 5 from both sides to get $7x^2 - 2x - 5 \leqslant 0$.
  2. Find the roots: Solve the corresponding quadratic equation (set the expression equal to zero) to find the roots. These roots are the points where the parabola intersects the x-axis. We can use factoring, the quadratic formula, or completing the square to find these roots. The quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, is a powerful tool for finding roots, especially when factoring is not straightforward.
  3. Determine the intervals: The roots divide the number line into intervals. We need to test a value from each interval in the original inequality to determine whether the inequality holds true in that interval. These intervals represent the potential solution sets for the inequality. For example, if the roots are x = 1 and x = 2, the intervals would be $(-\infty, 1)$, $(1, 2)$, and $(2, \infty)$.
  4. Write the solution: Based on the test values, identify the intervals that satisfy the inequality. The solution will be a union of these intervals. Remember to include the roots in the solution if the inequality is non-strict ($\leqslant$ or $\geqslant$), as they represent points where the expression equals zero.

Understanding these fundamental concepts and steps is crucial for tackling quadratic inequalities effectively. In the following sections, we will apply these principles to solve our specific problem and illustrate each step in detail.

Step 1: Rearranging the Inequality

The first crucial step in solving the given quadratic inequality, $7x^2 - 2x leqslant{leqslant} 5$, is to rearrange it into a standard form. This involves moving all terms to one side of the inequality, leaving zero on the other side. This rearrangement is essential because it allows us to easily identify the coefficients of the quadratic expression and set up the problem for factoring or using the quadratic formula.

To achieve this, we subtract 5 from both sides of the inequality:

7x^2 - 2x - 5 ${leqslant}$ 5 - 5

This simplifies to:

7x^2 - 2x - 5 ${leqslant}$ 0

Now, the inequality is in the standard form $ax^2 + bx + c leqslant{leqslant} 0$, where a = 7, b = -2, and c = -5. Having the inequality in this form is vital for the next steps, as it allows us to focus on the quadratic expression and determine its roots. The roots, as mentioned earlier, are the points where the quadratic expression equals zero, and they play a critical role in identifying the intervals where the inequality holds true.

By rearranging the inequality, we have set the stage for finding the roots and subsequently solving the inequality. This initial step is not just a matter of algebraic manipulation; it's a fundamental aspect of the problem-solving strategy. It transforms the inequality into a form that is easier to analyze and work with. In the next step, we will delve into finding the roots of the quadratic equation $7x^2 - 2x - 5 = 0$, which will pave the way for determining the solution set of the inequality.

Step 2: Finding the Roots of the Quadratic Equation

With the inequality now in the standard form $7x^2 - 2x - 5 leqslant{leqslant} 0$, the next step is to find the roots of the corresponding quadratic equation $7x^2 - 2x - 5 = 0$. The roots are the values of x that make the quadratic expression equal to zero. These roots are crucial because they divide the number line into intervals, which we will later test to determine the solution set of the inequality. There are several methods to find the roots of a quadratic equation, including factoring, completing the square, and using the quadratic formula. In this case, we will use the factoring method and the quadratic formula to illustrate both approaches.

Factoring Method

First, let's attempt to factor the quadratic expression. Factoring involves expressing the quadratic as a product of two binomials. We look for two numbers that multiply to give the product of a and c (7 * -5 = -35) and add up to b (-2). The numbers -7 and 5 satisfy these conditions (-7 * 5 = -35 and -7 + 5 = -2). We can rewrite the middle term using these numbers:

7x2βˆ’7x+5xβˆ’5=07x^2 - 7x + 5x - 5 = 0

Now, we factor by grouping:

7x(xβˆ’1)+5(xβˆ’1)=07x(x - 1) + 5(x - 1) = 0

We can factor out the common term $(x - 1)$:

(7x+5)(xβˆ’1)=0(7x + 5)(x - 1) = 0

Setting each factor equal to zero gives us the roots:

7x+5=0β‡’x=βˆ’577x + 5 = 0 \Rightarrow x = -\frac{5}{7}

xβˆ’1=0β‡’x=1x - 1 = 0 \Rightarrow x = 1

So, the roots of the quadratic equation are $x = -\frac{5}{7}$ and $x = 1$. These are the critical points that divide the number line into intervals.

Quadratic Formula

If factoring is not straightforward or possible, the quadratic formula is a reliable alternative. The quadratic formula is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For our equation, a = 7, b = -2, and c = -5. Plugging these values into the quadratic formula, we get:

x=βˆ’(βˆ’2)Β±(βˆ’2)2βˆ’4(7)(βˆ’5)2(7)x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(7)(-5)}}{2(7)}

x=2Β±4+14014x = \frac{2 \pm \sqrt{4 + 140}}{14}

x=2Β±14414x = \frac{2 \pm \sqrt{144}}{14}

x=2Β±1214x = \frac{2 \pm 12}{14}

This gives us two possible values for x:

x=2+1214=1414=1x = \frac{2 + 12}{14} = \frac{14}{14} = 1

x=2βˆ’1214=βˆ’1014=βˆ’57x = \frac{2 - 12}{14} = \frac{-10}{14} = -\frac{5}{7}

As we can see, both the factoring method and the quadratic formula yield the same roots: $x = -\frac{5}{7}$ and $x = 1$. These roots are the key to determining the intervals where the inequality $7x^2 - 2x - 5 leqslant{leqslant} 0$ holds true. In the next step, we will use these roots to divide the number line into intervals and test values within each interval to find the solution set.

Step 3: Determining the Intervals and Testing Values

Now that we have found the roots of the quadratic equation $7x^2 - 2x - 5 = 0$ to be $x = -\frac{5}{7}$ and $x = 1$, the next crucial step is to determine the intervals these roots create on the number line. These roots act as dividing points, splitting the number line into distinct regions where the quadratic expression will either be positive, negative, or zero. We will then test values within each interval to see if they satisfy the original inequality $7x^2 - 2x - 5 leqslant{leqslant} 0$.

The roots $x = -\frac{5}{7}$ and $x = 1$ divide the number line into three intervals:

  1. (- \infty, -\frac{5}{7})$ : This interval represents all values of *x* less than $-\frac{5}{7}$.

  2. (-\frac{5}{7}, 1)$: This interval represents all values of *x* between $-\frac{5}{7}$ and 1.

  3. (1, \infty)$: This interval represents all values of *x* greater than 1.

To determine which of these intervals satisfy the inequality $7x^2 - 2x - 5 leqslant{leqslant} 0$, we need to test a value from each interval in the inequality. This involves substituting a test value from each interval into the quadratic expression and checking if the result is less than or equal to zero. Let's choose test values for each interval:

  1. Interval $(- \infty, -\frac{5}{7})$: Let's choose x = -1. Substitute x = -1 into the inequality:

    7(βˆ’1)2βˆ’2(βˆ’1)βˆ’5=7+2βˆ’5=47(-1)^2 - 2(-1) - 5 = 7 + 2 - 5 = 4

    Since 4 is not less than or equal to 0, this interval does not satisfy the inequality.
  2. Interval $(-\frac{5}{7}, 1)$: Let's choose x = 0. Substitute x = 0 into the inequality:

    7(0)2βˆ’2(0)βˆ’5=βˆ’57(0)^2 - 2(0) - 5 = -5

    Since -5 is less than or equal to 0, this interval satisfies the inequality.
  3. Interval $(1, \infty)$: Let's choose x = 2. Substitute x = 2 into the inequality:

    7(2)2βˆ’2(2)βˆ’5=28βˆ’4βˆ’5=197(2)^2 - 2(2) - 5 = 28 - 4 - 5 = 19

    Since 19 is not less than or equal to 0, this interval does not satisfy the inequality.

From these tests, we find that only the interval $(-\frac{5}{7}, 1)$ satisfies the inequality $7x^2 - 2x - 5 < 0$. However, our original inequality is $7x^2 - 2x - 5 leqslant{leqslant} 0$, which includes the condition where the expression equals zero. This means we must also consider the roots themselves, $x = -\frac{5}{7}$ and $x = 1$, as part of the solution.

In the next step, we will combine the interval that satisfies the inequality with the roots to write the complete solution set. This will give us the final answer to the problem, providing the range of x values that make the inequality true.

Step 4: Writing the Solution

Having tested the intervals and identified that the inequality $7x^2 - 2x - 5 leqslant{leqslant} 0$ is satisfied in the interval $(-\frac{5}{7}, 1)$, the final step is to write the complete solution. We need to consider that our original inequality includes the