Question

Represent the solution graphically. Check the solution algebraically. $$ 2 x^{2}+4 x=6 $$

Short answer

The solutions to the equation \(2x^2 + 4x - 6 = 0\) are \(x = 1\) and \(x = -2\). The solutions are represented on the graph by the x-intercepts of the parabolic function.

Step-by-step solution

Normalize the Equation

First, let’s rearrange the equation to the form \(ax^2 + bx + c = 0\). We achieve this by subtracting 6 from both sides: \(2x^2 + 4x - 6 = 0 \)

Identify the Coefficients a, b and c

The coefficients are the numbers that multiply the variables or the constant term in a polynomial. Those will be needed for the quadratic formula. In this equation, the coefficients are a = 2, b = 4 and c = -6.

Solve with the Quadratic Formula

The quadratic formula allows us to solve any quadratic equation. Given \(ax^2 + bx + c = 0\), roots can be found using the formula \(x = [-b ± sqrt(b^2 - 4ac) ]/ 2a\). Substituting the values we get: \( x = [-4 ± sqrt((4)^2 - 4*2*(-6))]/(2*2)\) which simplifies to \(x = [-4 ± sqrt(16 + 48)]/4 = [-4 ± sqrt(64)]/4 = [-4 ± 8 ]/4 \). This gives two solutions \(x1 = 1\) and \(x2 = -2\).

Graph the Quadratic Function

Draw the x and y axes and mark the solutions \(x = 1\) and \(x = -2\) on the x-axis. The equation represents a parabolic curve opening upwards (since \(a > 0\)). The vertex lies between the roots.

Check the Solution Algebraically

The solutions will be checked by substituting them back into the original equation. Plugging \(x = 1\) back into the original equation we get: \(2*(1)^2 + 4*1 - 6 = 0\). For \(x = -2\), we get: \(2*(-2)^2 + 4*(-2) - 6 = 0\). As both sides of the equations are equal, our solutions are correct.

Key concepts

Solving Quadratic Equations

Understanding how to solve quadratic equations is fundamental in algebra. A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients and \( x \) represents an unknown variable. The quadratic formula, \( x = [-b \pm \sqrt{b^2 - 4ac} ]/ 2a \), is a powerful tool for finding the roots of any quadratic equation.

When applying the quadratic formula, it's crucial to first ensure the equation is normalized, which means it should be ordered and set equal to zero, as in the given exercise. Once you've rearranged the terms, identify your coefficients, which in our case are \( a = 2 \), \( b = 4 \), and \( c = -6 \). These values are then plugged into the quadratic formula to find the solutions.

Steps to Use the Quadratic Formula

• Rearrange the quadratic equation to the standard form.
• Identify and note down the coefficients \( a \), \( b \), and \( c \).
• Substitute the coefficients into the quadratic formula and simplify.
• Solve the resulting equation to find the value of \( x \).

Solving our equation with these steps yielded two solutions, \( x_1 = 1 \) and \( x_2 = -2 \), which shows the quadratic formula's efficacy in resolving quadratic equations.

Graphing Parabolas

Parabolas are the graphical representations of quadratic functions, characterized by their curved 'U' shape. When graphing a parabola for the equation \( ax^2 + bx + c = 0 \), it's important to find the roots or x-intercepts, which are the points where the parabola crosses the x-axis. These are the points we found earlier using the quadratic formula.

To graph the parabola, start by plotting the x-intercepts on a coordinate plane, then determine the direction in which the parabola opens. If \( a > 0 \), as it is in our exercise, the parabola opens upward. Conversely, if \( a < 0 \), it would open downward. The vertex of the parabola lies exactly midway between the x-intercepts and is the maximum or minimum point of the curve, depending on the direction the parabola opens.

Noteworthy Points When Graphing a Parabola

• Identify the x-intercepts using the roots from the quadratic formula.
• Determine the opening direction of the parabola based on the coefficient \( a \).
• Find the vertex, which is the peak or trough of the parabola and lies midway between the roots.
• Sketch the symmetric U-shaped curve through these points.

By applying these principles, students can visualise quadratic equations, aiding in a deeper conceptual understanding.

Quadratic Coefficients

The coefficients in a quadratic equation \( ax^2 + bx + c = 0 \) tell us a lot about the nature of its graph and solutions. The coefficient \( a \) determines the direction in which the parabola opens, while \( b \) affects the position of the parabola along the x-axis, and \( c \) represents the y-intercept, where the parabola crosses the y-axis.

The discriminant, which is part of the quadratic formula—specifically the expression under the square root, \( b^2 - 4ac \)—provides information about the nature of the roots. If the discriminant is positive, the quadratic equation has two distinct real roots. If it is zero, there is one real root, and if negative, there are no real roots, only complex ones.

Understanding the Impact of Coefficients

• The coefficient \( a \) influences the direction and width of the parabola.
• The coefficient \( b \) can affect the horizontal placement of the vertex.
• The coefficient \( c \) gives the y-intercept and indicates where the parabola crosses the y-axis.
• The discriminant reveals how many real roots the equation has.

In our exercise, the positive value of \( a \) led to an upward-opening parabola, and the positive discriminant resulted in two real, distinct roots. By understanding coefficients, students can predict the nature of parabolic graphs and the solutions to quadratic equations.