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Magazine Chapter 11: Problem 30
Use the quadratic formula to solve each equation. These equations have real solutions and complex, but not real, solutions. See Examples 1 through 4. $$ \frac{x^{2}}{2}-3=-\frac{9}{2} x $$
Short Answer
Expert verified
The solutions are \( x = \frac{-9 + \sqrt{105}}{2} \) and \( x = \frac{-9 - \sqrt{105}}{2} \).
Step by step solution
01
Rearrange the Equation
To use the quadratic formula, we first need to rewrite the given equation in standard quadratic form: \[ax^2 + bx + c = 0\]The given equation is \[\frac{x^2}{2} - 3 = -\frac{9}{2}x\]First, add \( \frac{9}{2}x \) to both sides to obtain:\[\frac{x^2}{2} + \frac{9}{2}x - 3 = 0\]
02
Multiply to Clear Fractions
To eliminate fractions, multiply the entire equation by 2 (the denominator of the fractions): \[ 2 \times \left( \frac{x^2}{2} + \frac{9}{2}x - 3 \right) = 0 \]This simplifies to: \[ x^2 + 9x - 6 = 0 \]
03
Identify Coefficients for the Quadratic Formula
Now that the equation is in standard form, identify the coefficients: - \(a = 1\) (coefficient of \(x^2\)),- \(b = 9\) (coefficient of \(x\)),- \(c = -6\) (constant term).
04
Apply the Quadratic Formula
The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the values of \(a\), \(b\), and \(c\) into the formula:\[ x = \frac{-9 \pm \sqrt{(9)^2 - 4(1)(-6)}}{2(1)} \]
05
Calculate the Discriminant
First, calculate the discriminant: \(b^2 - 4ac\).\[ (9)^2 - 4(1)(-6) = 81 + 24 = 105 \]The discriminant is 105, which is positive, indicating two real solutions.
06
Solve for the Roots
Now solve for \(x\) using the quadratic formula with the calculated discriminant:\[ x = \frac{-9 \pm \sqrt{105}}{2} \]This gives two solutions:\[ x = \frac{-9 + \sqrt{105}}{2} \]and\[ x = \frac{-9 - \sqrt{105}}{2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
Quadratic equations are fundamental expressions in algebra that involve a variable raised to the second power. They typically appear in the form \( ax^2 + bx + c = 0 \), where:
- \(a\), \(b\), and \(c\) are constants, with \(a eq 0\) to ensure the presence of the \(x^2\) term.
- The term \(ax^2\) is known as the quadratic term, \(bx\) is the linear term, and \(c\) is the constant term.
Discriminant
The discriminant in the context of a quadratic equation \( ax^2 + bx + c = 0 \) is expressed as \( b^2 - 4ac \). This value provides critical insight into the nature of the solutions:
- If \( b^2 - 4ac > 0 \), the quadratic equation has two distinct real solutions. This means that the parabola defined by the quadratic equation intersects the x-axis at two points.
- If \( b^2 - 4ac = 0 \), there is one real solution, often referred to as a repeated or double root. The parabola will touch the x-axis at one point.
- If \( b^2 - 4ac < 0 \), the solutions are complex and not real, indicating that the parabola does not intersect the x-axis at any point.
Complex Solutions
Complex solutions arise in quadratic equations when the discriminant is negative. When this happens, the solutions involve imaginary numbers, which are expressed as \( a + bi \), where \( i \) is the imaginary unit satisfying \( i^2 = -1 \).
- Complex solutions occur in conjugate pairs such as \( a + bi \) and \( a - bi \).
- These solutions do not correspond to points where a parabola crosses the real x-axis, but they are crucial in fields like engineering and physics where systems exhibit oscillations or resonance.
Standard Form
The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). This straightforward arrangement makes it easier to apply methods like factoring, completing the square, or using the quadratic formula.
- To convert an equation to standard form, all terms are brought to one side of the equation set to zero.
- Fractions within the equation are often cleared by multiplying through by common denominators to avoid unnecessary complexity when identifying coefficients \(a\), \(b\), and \(c\).
- This form allows for a systematic approach to solving quadratic equations by identifying the coefficients and plugging them into the quadratic formula.
