Problem 21 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ 2-2 x=3 x^{2} $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{{-1 + \sqrt{7}}}{3}\) and \(x = \frac{{-1 - \sqrt{7}}}{3}\).

Step 1 - Set the equation to standard form

The quadratic equation should be in the form $$ax^2 + bx + c = 0$$. Rearrange the given equation accordingly: $$2 - 2x = 3x^2$$ Subtract 2 from both sides:$$-2x = 3x^2 - 2$$Next, add 2x to both sides: $$0 = 3x^2 + 2x - 2$$. Thus, the standard form is: $$3x^2 + 2x - 2 = 0.$$

Step 2 - Identify coefficients a, b, and c

From the standard form equation, identify the coefficients: $$a = 3, b = 2, c = -2.$$

Step 3 - Use the quadratic formula

The quadratic formula is $$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$$. Plug in the identified coefficients: $$a = 3, b = 2, c = -2.$$

Step 4 - Calculate the discriminant

Calculate the discriminant $$\Delta = b^2 - 4ac$$. Substitute the values: $$\Delta = 2^2 - 4(3)(-2)$$ Simplify: $$\Delta = 4 + 24 = 28.$$

Step 5 - Compute the solutions

Now plug in the discriminant and the coefficients into the quadratic formula: $$x = \frac{{-2 \pm \sqrt{28}}}{2 \times 3}.$$ Simplify the square root and the denominator: $$x = \frac{{-2 \pm 2\sqrt{7}}}{6}$$ Divide the numerator by 6: $$x = \frac{{-1 \pm \sqrt{7}}}{3}.$$ The solutions are: $$x = \frac{{-1 + \sqrt{7}}}{3}$$ and $$x = \frac{{-1 - \sqrt{7}}}{3}.$$

Key Concepts

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations. It can handle any quadratic equation, even when factoring is difficult or impossible. The formula is expressed as follows: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\). Here, \(a\), \(b\), and \(c\) are coefficients from your quadratic equation in standard form \(ax^2 + bx + c = 0\). This formula gives the solutions, or roots, of the equation by considering the coefficients and the discriminant. The quadratic formula ensures you can always find solutions provided you follow the correct steps.

discriminant

The discriminant is a crucial part of the quadratic formula. It's located under the square root symbol and is calculated with the expression \( \Delta = b^2 - 4ac \). The discriminant determines the nature of the solutions:

• If \( \Delta > 0 \), there are two distinct real solutions.
• If \( \Delta = 0 \), there is exactly one real solution (a repeated root).
• If \( \Delta < 0 \), there are no real solutions (the solutions are complex or imaginary).

This value tells you not only about the number of solutions but also their type, helping you understand the equation better before solving it entirely.

standard form of quadratic equation

When solving quadratic equations, it's essential first to rewrite them in standard form. The standard form of a quadratic equation is: \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants. This format ensures you can apply the quadratic formula effectively. To transform any quadratic equation into standard form, you may need to rearrange the terms, combine like terms, or move them across the equality sign. For example, if you start with \(2 - 2x = 3x^2\), you would rearrange it to \(3x^2 + 2x - 2 = 0\) by moving all terms to one side of the equation. Once in standard form, identifying the coefficients \(a\), \(b\), and \(c\) is straightforward, which are crucial for applying the quadratic formula.