Chapter 10: Problem 21
Solve equation. Approximate the solutions to the nearest hundredth. \(x^{2}=x+7\)
Short Answer
Expert verified
The solutions are approximately \(x \approx 3.20\) and \(x \approx -2.20\).
Step by step solution
01
Rearrange the Equation
Start by rearranging the equation to set it to zero on one side. The given equation is \(x^2 = x + 7\). Subtract \(x + 7\) from both sides to get: \(x^2 - x - 7 = 0\). This is the quadratic equation in the standard form \(ax^2 + bx + c = 0\).
02
Identify Coefficients
Identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation \(x^2 - x - 7 = 0\). Here, \(a = 1\), \(b = -1\), and \(c = -7\).
03
Apply the Quadratic Formula
Use the quadratic formula to find the solutions: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values of \(a\), \(b\), and \(c\) into the formula: \(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}\).
04
Simplify the Expression
Simplify the expression under the square root and calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4 \times 1 \times (-7) = 1 + 28 = 29\). So, the expression becomes \(x = \frac{1 \pm \sqrt{29}}{2}\).
05
Calculate the Solutions
Calculate the two solutions. For \(x = \frac{1 + \sqrt{29}}{2}\), use a calculator to approximate \(\sqrt{29} \approx 5.39\), and thus \(x \approx \frac{1 + 5.39}{2} = 3.20\). For \(x = \frac{1 - \sqrt{29}}{2}\), compute \(x \approx \frac{1 - 5.39}{2} = -2.20\).
06
Approximations
Both solutions are approximated to the nearest hundredth, which results in the solutions \(x \approx 3.20\) and \(x \approx -2.20\).
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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 fundamental tool used for solving quadratic equations of the form \(ax^2 + bx + c = 0\). Quadratic equations are second-degree polynomials, meaning the highest power of the variable \(x\) is 2. The quadratic formula provides a reliable method for finding the roots or solutions of any quadratic equation, no matter how complex it might seem. The formula is given by:
- \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Discriminant
The discriminant is a key part of the quadratic formula and is represented by \(b^2 - 4ac\). It helps in determining the nature and the number of solutions you can expect from a quadratic equation. Depending on the value of the discriminant, you can quickly gauge how many solutions there are and what type they will be:
- If the discriminant is greater than zero (\(b^2 - 4ac > 0\)), the quadratic equation has two distinct real roots.
- If the discriminant is equal to zero (\(b^2 - 4ac = 0\)), there is exactly one real root, meaning the parabola touches the x-axis at one point.
- When the discriminant is less than zero (\(b^2 - 4ac < 0\)), no real roots exist and the solutions are complex or imaginary numbers, meaning the equation never actually crosses the x-axis.
Coefficients
Coefficients are integral parts of quadratic equations and play a crucial role in using the quadratic formula. In a quadratic equation of the form \(ax^2 + bx + c = 0\):
- \(a\) is the coefficient of \(x^2\) and is known as the quadratic coefficient. It determines the opening of the parabola (upward or downward).
- \(b\) is the coefficient of \(x\), called the linear coefficient, affecting the vertex and axis of symmetry of the parabola.
- \(c\) is the constant term, impacting the vertical position or the y-intercept of the parabola.