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Chapter 1: Problem 5

Determine whether the given value is a solution of the equation. $$m^{2}+m-\frac{5}{16}=0 ; m=\frac{1}{4}$$

Short Answer

Expert verified
Yes, \( m=\frac{1}{4} \) is a solution of the equation.

Step by step solution

01

Substitute the given value into the equation

We are asked to determine if \( m = \frac{1}{4} \) is a solution to the equation \( m^2 + m - \frac{5}{16} = 0 \). Start by substituting \( m = \frac{1}{4} \) into the equation.
02

Calculate \( m^2 \)

First, calculate \( m^2 \) when \( m = \frac{1}{4} \). \( \left( \frac{1}{4} \right)^2 = \frac{1}{16} \).
03

Calculate the left side of the equation

With \( m = \frac{1}{4} \), calculate \( m^2 + m - \frac{5}{16} \).Substitute the values:\( \frac{1}{16} + \frac{1}{4} - \frac{5}{16} \).
04

Add \( \frac{1}{16} \) and \( \frac{1}{4} \)

Convert \( \frac{1}{4} \) to \( \frac{4}{16} \) to add: \( \frac{1}{16} + \frac{4}{16} = \frac{5}{16} \).
05

Subtract \( \frac{5}{16} \) from the result

Now subtract \( \frac{5}{16} \) from the result: \( \frac{5}{16} - \frac{5}{16} = 0 \).
06

Verify the solution

Since substituting \( m = \frac{1}{4} \) into the equation results in 0, which matches the right side of the equation, \( m = \frac{1}{4} \) is a solution.

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Key Concepts

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

Solution Verification
To confirm if a given value is a solution to an equation, we must substitute it into the equation and see if it satisfies the equation's statement. In this case, we have a quadratic equation: \[ m^2 + m - \frac{5}{16} = 0 \] The process involves checking whether substituting \( m = \frac{1}{4} \) results in a true equation.
  • First, calculate by substituting \( m = \frac{1}{4} \) into \( m^2 \), \( m \), and the constant term in the equation.
  • Next, evaluate whether the left-hand side equals the right-hand side, which is zero in this case.
When no discrepancy is seen and the statement holds, the value is verified as a solution for the equation.
Substitution Method
The substitution method is a way to determine if a specific value can satisfy an equation. It involves replacing variables with numbers and calculating the result. To use this method in our exercise:
  • Replace \( m \) with \( \frac{1}{4} \) wherever \( m \) appears in the equation: \[ \left( \frac{1}{4} \right)^2 + \frac{1}{4} - \frac{5}{16} = 0 \]
  • Perform the necessary arithmetic such as squaring, addition, and subtraction.
This method is extremely useful for checking solutions because it allows you to see directly if both sides of an equation remain equal after substitution.
Solving Equations
Solving equations like a quadratic can seem challenging, but with a structured approach, it becomes simpler. A quadratic equation typically takes the form:\[ ax^2 + bx + c = 0 \] In our exercise, we've already been provided with a possible solution to verify. However, finding these solutions manually usually involves:
  • Factoring the quadratic, if possible.
  • Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  • Employing the method of completing the square.
After finding a potential solution, using substitution to verify is an excellent way to ensure correctness, just as shown in the exercise.

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Most popular questions from this chapter

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$x=y^{3}-1$$

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the interval(s) on a number line. $$|x-1| \leq \frac{1}{2}$$

Use a graphing utility to graph the equations and to approximate the \(x\) -intercepts. In approximating the \(x\) -intercepts, use a "solve" key or a sufficiently magnified view to ensure that the values you give are correct in the first three decimal places. Remark: None of the \(x\) -intercepts for these four equations can be obtained using factoring techniques.) $$y=8 x^{3}-6 x-1$$

This exercise outlines a proof of the fact that two nonvertical lines with slopes \(m_{1}\) and \(m_{2}\) are perpendicular if and only if \(m_{1} m_{2}=-1 .\) In the following figure, we've assumed that our two nonvertical lines \(y=m_{1} x\) and \(y=m_{2} x\) intersect at the origin. [If they did not intersect there, we could just as well work with lines parallel to these that do intersect at \((0,0),\) recalling that parallel lines have the same slope.] The proof relies on the following geometric fact: \(\overline{O A} \perp \overline{O B} \quad\) if and only if \(\quad(O A)^{2}+(O B)^{2}=(A B)^{2}\). (a) Verify that the coordinates of \(A\) and \(B\) are \(A\left(1, m_{1}\right)\) and \(B\left(1, m_{2}\right)\) (b) Show that $$\begin{aligned}O A^{2} &=1+m_{1}^{2} \\\O B^{2} &=1+m_{2}^{2} \\\A B^{2} &=m_{1}^{2}-2 m_{1} m_{2}+m_{2}^{2} \end{aligned}$$ (c) Use part (b) to show that the equation $$O A^{2}+O B^{2}=A B^{2}$$ is equivalent to \(m_{1} m_{2}=-1\) (GRAPH CAN'T COPY)

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the interval(s) on a number line. $$|x|>1$$
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