Problem 40 - Algebra 2

Question

Solve each equation by factoring, by taking square roots, or by graphing. If necessary, round your answer to the nearest hundredth. $$ x^{2}+2 x=6-6 x $$

Step-by-Step Solution

Answer

The solutions to the equation are approximately \( x \approx 0.69 \) and \( x \approx -8.69 \), rounded to the nearest hundredth.

Step 1: Equate the equation to zero

In order to solve the given quadratic equation by factoring or by taking square roots, first rearrange the terms to set the equation to zero. Do this by moving all terms to one side of the equation. \[ x^{2} + 2x - 6 + 6x = 0 \] Combine like terms. \[ x^{2} + 8x - 6 = 0 \]

Step 2: Factor the quadratic equation

The equation does not factor nicely into integer roots. In such cases, either complete the square or use the quadratic formula. Here, we will proceed to use the quadratic formula.

Step 3: Apply the quadratic formula

For a quadratic equation in the form \( ax^{2} + bx + c = 0 \), the solutions for \( x \) can be found using the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^{2}-4ac}}}}{{2a}} \). Substitute \( a = 1 \), \( b = 8 \), and \( c = -6 \) into the formula. \[ x = \frac{{-8 \pm \sqrt{{8^{2}-4(1)(-6)}}}}{{2(1)}} \] \[ x = \frac{{-8 \pm \sqrt{{64+24}}}}{{2}} \] \[ x = \frac{{-8 \pm \sqrt{{88}}}}{{2}} \]

Step 4: Simplify and find the roots

Simplify the expression under the square root and find two possible solutions for \( x \). \[ x = \frac{{-8 \pm \sqrt{{88}}}}{{2}} \] Simplify \( \sqrt{{88}} \) as \( 2\sqrt{{22}} \). \[ x = \frac{{-8 \pm 2\sqrt{{22}}}}{{2}} \] Divide each term by 2. \[ x = -4 \pm \sqrt{{22}} \]

Step 5: Approximate to the nearest hundredth if necessary

We can approximate \( \sqrt{{22}} \) to get the solutions to the nearest hundredth. \[ x \approx -4 \pm 4.69 \] So we have two solutions: \[ x \approx -4 + 4.69 = 0.69 \]and\[ x \approx -4 - 4.69 = -8.69 \]

Key Concepts

Factoring Quadratics

When you encounter a quadratic equation, factoring is one of the simplest methods to find its solutions, provided that the equation can be easily factored into two binomials. Factoring involves expressing the quadratic equation in the form \(ax^2 + bx + c = 0\) as a product of two binomials. For example, if you have \(x^2 + 5x + 6 = 0\), it factors into \(x + 2\) and \(x + 3\), so the solutions are \(x = -2\) and \(x = -3\).

However, not all quadratics will factor neatly. If the equation doesn't factor into integers, as seen in our original exercise, you'd need to consider other methods such as completing the square or using the quadratic formula. Keep in mind the importance of setting the equation equal to zero before attempting to factor, and combine like terms to simplify the equation as necessary.

Quadratic Formula

The quadratic formula provides a solution for any quadratic equation of the form \(ax^2 + bx + c = 0\) when factoring is not possible or practical. The formula is given by \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\). The term \(\pm\) means that there will be two solutions: one where you add the square root term, and the other where you subtract it. These solutions can sometimes be irrational numbers, requiring you to round them to the nearest hundredth or other decimal place, as we did in our original exercise.

Using the quadratic formula can seem daunting at first, but with practice, you'll recognize it as a reliable tool for solving any quadratic equation. Just remember to substitute the appropriate values for \(a\), \(b\), and \(c\), then simplify to find the roots.

Completing the Square

Completing the square is another method for solving quadratic equations, particularly when they cannot be factored easily. This technique transforms the quadratic into a perfect square trinomial, which can then be solved by taking the square root of both sides.

To complete the square, start by arranging the quadratic equation so that the \(x^2\) and \(x\) terms are on one side and the constant is on the other. Then add the square of half the coefficient of \(x\) to both sides. For example, for an equation like \(x^2 - 4x = 21\), you'd add \((\frac{{-4}}{2})^2 = 4\) to both sides, resulting in \(x^2 - 4x + 4 = 25\). The left side is now a perfect square \((x - 2)^2\) and easy to solve.

Even when not used as the primary solution method, completing the square is integral to understanding the derivation of the quadratic formula and provides a deeper comprehension of quadratic equations.