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Magazine Chapter 3: Problem 38
Solve each quadratic equation (a) graphically, (b) numerically, and (c) symbolically. Express graphical and numerical solutions to the nearest tenth when appropriate. $$ 2 x^{2}+5 x-3=0 $$
Short Answer
Expert verified
The solutions are \(x \approx -3.0\) and \(x \approx 0.5\).
Step by step solution
01
Convert to Standard Form
The given quadratic equation is \(2x^2 + 5x - 3 = 0\). This equation is already in the standard form \(ax^2 + bx + c = 0\), where \(a = 2\), \(b = 5\), and \(c = -3\).
02
Graphical Solution
To solve the equation graphically, plot the quadratic function \(y = 2x^2 + 5x - 3\) on a graph. Identify the points where the parabola intersects the x-axis. These points are the roots of the equation. After plotting, you will observe that the parabola crosses the x-axis at approximately \((-3.0, 0)\) and \(0.5, 0)\). Thus, the graphical solutions are \(x \approx -3.0\) and \(x \approx 0.5\).
03
Numerical Solution using Quadratic Formula
Apply the quadratic formula, which is \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], to find the roots. Substitute \(a = 2\), \(b = 5\), and \(c = -3\) into the formula:\[x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}\]Calculate the discriminant:\[b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot (-3) = 25 + 24 = 49\]Since the discriminant is a perfect square, compute the solutions:\[x = \frac{-5 \pm \sqrt{49}}{4}\]\[x = \frac{-5 \pm 7}{4}\]The solutions are \(x = \frac{-5 + 7}{4} = 0.5\) and \(x = \frac{-5 - 7}{4} = -3.0\).
04
Verify Solutions Symbolically
Substitute \(x = 0.5\) and \(x = -3\) back into the original equation and verify:For \(x = 0.5\):\[2(0.5)^2 + 5(0.5) - 3 = 2(0.25) + 2.5 - 3 = 0.5 + 2.5 - 3 = 0\]For \(x = -3\):\[2(-3)^2 + 5(-3) - 3 = 18 - 15 - 3 = 0\]Both solutions satisfy the equation, confirming the solutions are correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphical Solutions
Graphical solutions involve using a graph to find the roots of a quadratic equation. For the given equation, \(2x^2 + 5x - 3 = 0\), the corresponding function \(y = 2x^2 + 5x - 3\) forms a parabola when plotted on a coordinate plane. To find the roots graphically, you're looking for points where this parabola intersects the x-axis.
- The x-coordinate of these intersection points represents the roots of the equation.
- In this example, the parabola crosses the x-axis at approximately \(x \approx -3.0\) and \(x \approx 0.5\).
Numerical Solutions
Numerical solutions involve precise calculations to determine the roots. Though graphing provides a general idea where the solutions are, numerical methods give exact values. In our solved exercise, we used the quadratic formula for a numerical approach.The quadratic formula is an effective tool for numerical solutions:
- Substitute the coefficients \(a\), \(b\), and \(c\) from the standard form equation \(ax^2 + bx + c = 0\) into the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- Perform precise calculations on the expression under the square root, known as the discriminant \(b^2 - 4ac\).
Quadratic Formula
The quadratic formula provides a direct way to find solutions of any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is essential because it avoids the need to factor which might not be feasible for all equations. Its forms are:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Key steps to using the quadratic formula:
- Calculate the discriminant \(b^2 - 4ac\). The sign of the discriminant (positive, zero, or negative) reveals how many real solutions exist.
- Substitute into the formula to find each root. The \(\pm\) symbol indicates that there are generally two solutions, except when the discriminant equals zero.
- Solve for \(x\) to get the two possible roots.
Standard Form
The standard form of a quadratic equation is expressed as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. It's called standard form because of its structured format which simplifies the process of manipulating the equation.
- In this format, you can easily identify the coefficients needed for methods like the quadratic formula or factoring.
- The precise understanding of \(a\) as the leading coefficient helps determine the direction (upward or downward) of the parabola in graphical solutions.
