Chapter 6: Problem 10
First use the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real solutions. Then solve the equation. $$2 x^{2}-6 x=-1$$
Short Answer
Expert verified
The equation has two real solutions: \( x = \frac{3 + \sqrt{7}}{2} \) and \( x = \frac{3 - \sqrt{7}}{2} \).
Step by step solution
01
Rearrange the Equation
First, bring the equation into the standard quadratic form of \( ax^2 + bx + c = 0 \). Start with the given equation \( 2x^2 - 6x = -1 \). Add 1 to both sides to get: \( 2x^2 - 6x + 1 = 0 \). Now, the equation fits the quadratic form where \( a = 2 \), \( b = -6 \), and \( c = 1 \).
02
Calculate the Discriminant
The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \), denoted as \( \Delta \), is given by the formula \( \Delta = b^2 - 4ac \). Substituting \( a = 2 \), \( b = -6 \), and \( c = 1 \) into the formula gives: \( \Delta = (-6)^2 - 4 \times 2 \times 1 = 36 - 8 = 28 \).
03
Interpret the Discriminant
The value of the discriminant is \( \Delta = 28 \). Because \( \Delta > 0 \), the quadratic equation \( 2x^2 - 6x + 1 = 0 \) has two distinct real solutions.
04
Solve the Quadratic Equation
Since the discriminant indicates two real solutions, use the quadratic formula to find them. The formula is \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \). Substitute \( a = 2 \), \( b = -6 \), and \( \Delta = 28 \) into the formula to find the solutions: \[ x = \frac{-(-6) \pm \sqrt{28}}{2 \times 2} = \frac{6 \pm \sqrt{28}}{4} \]. Further simplifying \( \sqrt{28} \) as \( 2\sqrt{7} \) gives: \[ x = \frac{6 \pm 2\sqrt{7}}{4} = \frac{3 \pm \sqrt{7}}{2} \]. Therefore, the two solutions are \( x = \frac{3 + \sqrt{7}}{2} \) and \( x = \frac{3 - \sqrt{7}}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Discriminant
The discriminant is a crucial component in solving quadratic equations. It is the part of the quadratic formula under the square root, expressed as \( b^2 - 4ac \). This value, often called \( \Delta \), helps us determine the nature of the roots of a quadratic equation. By examining \( \Delta \), you can instantly know if the equation has real or complex solutions.
When solving the quadratic equation \( 2x^2 - 6x + 1 = 0 \), we identified a, b, and c as \( 2, -6, \) and \( 1 \) respectively. Plugging these values into the discriminant formula gave us \( \Delta = 28 \). This is greater than zero, indicating that there are two distinct real solutions.
The relationship between the discriminant and the nature of the roots is straightforward:
When solving the quadratic equation \( 2x^2 - 6x + 1 = 0 \), we identified a, b, and c as \( 2, -6, \) and \( 1 \) respectively. Plugging these values into the discriminant formula gave us \( \Delta = 28 \). This is greater than zero, indicating that there are two distinct real solutions.
The relationship between the discriminant and the nature of the roots is straightforward:
- If \( \Delta > 0 \), the equation has two distinct real solutions.
- If \( \Delta = 0 \), the equation has exactly one real solution, with a repeated root or multiplicity of two.
- If \( \Delta < 0 \), the equation has two nonreal, complex solutions.
Real Solutions
Real solutions of a quadratic equation refer to the values of \( x \) that satisfy the equation and are real numbers (not involving imaginary units). In the given problem, the discriminant \( \Delta = 28 \) indicated that our equation \( 2x^2 - 6x + 1 = 0 \) possesses two distinct real solutions.
This means when we solve the equation, both solutions will be real numbers, which are easy to plot on a standard number line.
Understanding the nature of solutions helps in predicting the type of graph the quadratic equation will produce. For distinct real solutions:
This means when we solve the equation, both solutions will be real numbers, which are easy to plot on a standard number line.
Understanding the nature of solutions helps in predicting the type of graph the quadratic equation will produce. For distinct real solutions:
- The graph of the equation will intersect the x-axis at two separate points, corresponding to the two real solutions.
- This indicates a parabola opening upwards or downwards, depending on the sign of coefficient \( a \).
Quadratic Formula
The quadratic formula provides a reliable method for finding the roots of a quadratic equation. This formula is especially useful when factoring is tricky or the solutions are non-integer or irrational.
The formula itself is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting values from our problem \( a = 2 \), \( b = -6 \), and \( \Delta = 28 \) (from our discriminant calculation), gives us:\[ x = \frac{6 \pm \sqrt{28}}{4} \]
This simplifies to:\[x = \frac{3 \pm \sqrt{7}}{2}\]resulting in two real solutions: \( x = \frac{3 + \sqrt{7}}{2} \) and \( x = \frac{3 - \sqrt{7}}{2} \).
Steps using the quadratic formula:
The formula itself is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting values from our problem \( a = 2 \), \( b = -6 \), and \( \Delta = 28 \) (from our discriminant calculation), gives us:\[ x = \frac{6 \pm \sqrt{28}}{4} \]
This simplifies to:\[x = \frac{3 \pm \sqrt{7}}{2}\]resulting in two real solutions: \( x = \frac{3 + \sqrt{7}}{2} \) and \( x = \frac{3 - \sqrt{7}}{2} \).
Steps using the quadratic formula:
- Calculate the discriminant \( b^2 - 4ac \) for insight into the nature of the roots.
- Substitute the values of \( a \), \( b \), and the discriminant into the formula.
- Solve for \( x \) to find the roots which the equation equates to zero.