Question

The solution of a quadratic equation can be found by graphing each side separately and locating the points of intersection. You may wish to consult page 532 for help in approximating solutions. $$ -5 x^{2}+4 x=2 x^{2}-8 $$

Short answer

The solution to the quadratic equation is \(x = 2 \pm \sqrt{60}\).

Step-by-step solution

Simplify the Equation

Begin by simplifying the equation. Do this by moving all the terms to one side of the equation in order to have it equal to zero. The equation can be rewritten as: \((-5x^{2} + 4x) - (2x^{2} - 8) = 0\), which simplifies to \(-7x^{2} + 4x + 8 = 0\).

Apply the Quadratic Formula

The quadratic formula is given by \(-b \pm \sqrt{b^{2}-4ac} \over 2a\). In this case, the coefficients are \(a = -7\), \(b = 4\), and \(c = 8\). Substituting these into the quadratic formula yields:\n\(x = -4 \pm \sqrt{(4)^{2} -4(-7)(8)} \over 2(-7)\)

Solve for x

Computing the values within the square root first gives:\n \(x = -4 \pm \sqrt{16 + 224} \over -14\)\n This simplifies to: \n\\(x = -4 \pm \sqrt{240} \over -14\)\n Simplify further to get the corresponding values of x: \n\\(x = 2 \pm \sqrt{60}\)

Key concepts

Graphing Quadratic Equations

When it comes to visualizing the solutions to a quadratic equation, graphing is an excellent tool. The basic form of a quadratic equation is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. Graphically, the solutions to the equation represent the points where the parabola, defined by the quadratic equation, intersects the \(x\)-axis.

To graph a quadratic equation, you should first convert it into standard form if it’s not already. In our example, \( -5x^2 + 4x = 2x^2 - 8 \) becomes \( -7x^2 + 4x + 8 = 0 \) after rearranging the terms. Next, identify the coefficients \(a = -7\), \(b = 4\), and \(c = 8\), which will influence the shape and position of the parabola. A negative \(a\) value indicates the parabola opens downward.

Plotting this on a graph, we look for points where the curve crosses the \(x\)-axis; these intersections are the real solutions to the equation. If the parabola does not touch the \(x\)-axis, it means there are no real solutions. To aid in this process, establishing a table of values based on the quadratic equation helps in sketching a more accurate graph.

Quadratic Formula

The quadratic formula provides a foolproof method for finding the solutions of a quadratic equation, \(ax^2 + bx + c = 0\), regardless of whether they are real or complex numbers. The formula is \(x = \frac{-b \: \: \sqrt{b^2 - 4ac}}{2a}\). It directly provides both possible values for \(x\), known as the roots of the equation.

Applying the quadratic formula to our initial equation \( -5x^2 + 4x = 2x^2 - 8 \), after simplification to \( -7x^2 + 4x + 8 = 0 \), involves identifying \(a = -7\), \(b = 4\), and \(c = 8\). Plugging these into the formula gives two possible solutions for \(x\) after simplifying under the radical and dividing by \(2a\). It's essential to compute carefully, particularly under the square root, as this determines whether the solutions are real or complex.

For example, in our case, we have \(x = \frac{-4 \: \: \sqrt{16 + 4(-7)(8)}}{-14}\) which simplifies to \(x = \frac{-4 \: \: \sqrt{240}}{-14}\), yielding the solutions \(x = 2 \: \: \pm \sqrt{60}\). These specific steps, when followed correctly, ensure a clear path to finding the roots.

Solving Quadratic Equations

There are multiple methods to tackle solving quadratic equations, and choosing the right one depends on the specifics of the equation you're working with. An important step is to always start by arranging the equation in standard form, \(ax^2 + bx + c = 0\).

For equations that are easily factorable, factoring is the most straightforward approach. However, when factoring is not feasible, the quadratic formula is a reliable alternative that works in all cases, as it does not rely on the equation being factorable.

Another method is completing the square, which involves creating a perfect square trinomial from the equation and then solving it. This is an excellent way to understand the concept of quadratic equations deeply. Using the example \( -7x^2 + 4x + 8 = 0 \), we would look for the square both sides of the equation that makes the \(x^2\) term complete. Lastly, a graphical solution using a plot of the equation may also reveal the real roots, as it did in the exercise we discussed. It is worth noting, for complex solutions or when high accuracy is required, the quadratic formula usually offers the best solution.