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Chapter 10: Problem 27

Solve equation. Approximate the solutions to the nearest hundredth. \(4 m^{2}=4 m+19\)

Short Answer

Expert verified
The approximate solutions are \(m \approx 2.74\) and \(m \approx -1.74\).

Step by step solution

01

Set up the Quadratic Equation

The given equation is \(4m^2 = 4m + 19\). To solve it, we must first set this equation to zero. Start by subtracting \(4m + 19\) from both sides, resulting in: \(4m^2 - 4m - 19 = 0\). This is a quadratic equation of the form \(ax^2 + bx + c = 0\).
02

Calculate the Discriminant

The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(D = b^2 - 4ac\). For this equation, \(a = 4\), \(b = -4\), and \(c = -19\). Substitute these values into the formula to get \(D = (-4)^2 - 4(4)(-19) = 16 + 304 = 320\).
03

Use the Quadratic Formula

The roots of a quadratic equation can be found using the quadratic formula \(m = \frac{-b \pm \sqrt{D}}{2a}\). Substitute \(b = -4\), \(a = 4\), and \(D = 320\) into the formula to find the solutions: \(m = \frac{-(-4) \pm \sqrt{320}}{2 \times 4}\). Simplify: \(m = \frac{4 \pm \sqrt{320}}{8}\).
04

Calculate the Roots

First, calculate \(\sqrt{320}\). Since \(320 = 16 \times 20\), \(\sqrt{320} = \sqrt{16 \times 20} = 4\sqrt{20}\). Approximating \(\sqrt{20} \approx 4.47\), find \(\sqrt{320} \approx 4 \times 4.47 = 17.88\). Substitute back to find \(m\): \(m = \frac{4 \pm 17.88}{8}\). So, \(m_1 = \frac{4 + 17.88}{8}\) and \(m_2 = \frac{4 - 17.88}{8}\).
05

Approximate the Solutions

Calculate \(m_1\): \(m_1 = \frac{21.88}{8} \approx 2.735\), rounding to the nearest hundredth is \(m_1 \approx 2.74\). Calculate \(m_2\): \(m_2 = \frac{-13.88}{8} \approx -1.735\), rounding to the nearest hundredth is \(m_2 \approx -1.74\).

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Key Concepts

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

Quadratic Formula
The quadratic formula is a powerful tool used to find the solutions or roots of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula is expressed as:\[m = \frac{-b \pm \sqrt{D}}{2a}\]Here's a step-by-step breakdown of how this formula works:
  • The letter \( b \) in the formula represents the coefficient of the \( x \) term in the equation.
  • The constant \( c \) is the standalone number.
  • The quadratic coefficient \( a \) is the multiplier of the \( x^2 \) term.
  • To find the solutions \( m \) (which are referred to as roots), we substitute these coefficients \( a \), \( b \), and the discriminant \( D \) (discussed in the next section) into the quadratic formula.
This formula allows us to solve quadratic equations regardless of whether they have real or complex solutions. It becomes particularly handy when the equation doesn't factor neatly, or when the roots are not integers.
Discriminant
The discriminant is a key component in the quadratic formula. It is derived from the coefficients of a quadratic equation and is calculated as:\[D = b^2 - 4ac\]Here is why the discriminant is essential:
  • **Determines the number of real solutions:** If \( D > 0 \), the quadratic equation has two distinct real roots. If \( D = 0 \), there is exactly one real root. If \( D < 0 \), the equation does not have any real roots; instead, it has two complex roots.
  • The value of the discriminant gives insight into the nature of the roots without actually needing to calculate the roots themselves.
  • In the exercise, \( D = 320 \), which is positive, implying that the quadratic equation has two distinct real solutions.
The discriminant acts like a gatekeeper. It lets us know not only if solutions exist, but also how the graph of the quadratic equation will intersect the x-axis.
Roots of Equation
The roots of a quadratic equation are the solutions that satisfy the equation, typically represented as the values of \( x \) or, in this exercise, \( m \), that make the equation true. Using the discriminant and quadratic formula, we compute the roots as follows: \[m = \frac{-b \pm \sqrt{D}}{2a}\]Here's how this applies to our exercise:
  • Substitute \( a = 4 \), \( b = -4 \), and \( D = 320 \) to find the roots.
  • The roots were calculated as \( m_1 = \frac{21.88}{8} \approx 2.74 \) and \( m_2 = \frac{-13.88}{8} \approx -1.74 \).
  • These roots can be verified by substituting them back into the original equation to ensure they satisfy it.
Finding the roots is about determining where the equation equals zero, which graphically translates into the points where the parabola intersects the x-axis. Understanding these intersections helps visualize and solve real-world problems modeled by quadratic equations.

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Most popular questions from this chapter

The surface area \(S\) of a sphere is given by the formula \(S=4 \pi r^{2},\) where \(r\) is the radius of the sphere. An almanac lists the surface area of the Earth as \(196,938,800\) square miles. Assuming the Earth to be spherical, what is its radius to the nearest mile?

Determine the coordinates of the vertex of the graph of each function using the vertex formula. Then determine the \(x-\) and \(y\) -intercepts of the graph. Finally, plot several points and complete the graph. See Example \(9 .\) $$ f(x)=-2 x^{2}+8 x-10 $$

Use a graphing calculator to solve each equation. If an answer is not exact, round to the nearest hundredth. See Using Your Calculator: Solving Quadratic Equations Graphically. $$ x^{2}+x-6=0 $$

Find a number between 0 and 1 such that the difference of the number and its square is a maximum.

Use a graphing calculator to solve each equation. If an answer is not exact, round to the nearest hundredth. See Using Your Calculator: Solving Quadratic Equations Graphically. $$ 0.5 x^{2}-0.7 x-3=0 $$
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