Problem 48 - Contemporary Precalculus

Question

Solve the equation by any method. $$2 x^{2}=6 x+3$$

Step-by-Step Solution

Answer

Answer: The solutions for the quadratic equation are \(x = \frac{3+\sqrt{15}}{2}\) and \(x = \frac{3-\sqrt{15}}{2}\).

Step 1: Write down the quadratic formula

The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where \(a\), \(b\), and \(c\) are the coefficients of the equation \(ax^2 + bx + c = 0\).

Step 2: Identify the coefficients

In our equation, \(2x^2 = 6x + 3\), we can rewrite it as \(2x^2 - 6x - 3 = 0\). Now we can see that \(a = 2\), \(b = -6\), and \(c = -3\).

Step 3: Substitute the coefficients into the quadratic formula

From Step 2, we know the coefficients are \(a = 2\), \(b = -6\), and \(c = -3\). Now we can substitute them into the quadratic formula: $$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-3)}}{2(2)}$$

Step 4: Simplify the expression

Let's first simplify the expression inside the square root: $$(-6)^2 - 4(2)(-3) = 36 + 24 = 60$$ Now we can substitute this value back into the formula and perform the calculations: $$x = \frac{6 \pm \sqrt{60}}{4}$$

Step 5: Simplify the square root (if possible)

In this case, we can simplify \(\sqrt{60}\) as follows: \(\sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15}\). So the final equation turns into: $$x = \frac{6 \pm 2\sqrt{15}}{4}$$

Step 6: Simplify the fractions

Now we can simplify the fractions by dividing by a common factor, which is 2 in this case: $$x = \frac{6 \pm 2\sqrt{15}}{4} = \frac{3 \pm \sqrt{15}}{2}$$ The two solutions for the given quadratic equation are: $$x = \frac{3 + \sqrt{15}}{2}$$ and $$x = \frac{3 - \sqrt{15}}{2}$$

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions to quadratic equations. These are equations in the form of \( ax^2 + bx + c = 0 \). The formula is defined as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This means, once you identify the coefficients \(a\), \(b\), and \(c\), you can substitute them into this formula to find the possible values of \(x\). It works universally for all quadratic equations, regardless of how they are formed or how complex they might look initially. Understanding this formula is key to solving quadratics with ease.
The \( \pm \) symbol signifies that the formula provides two solutions, one for the positive square root and one for the negative, reflecting the possibilities in a quadratic's graph, which forms a parabola and usually intersects the x-axis at up to two distinct points.

Coefficient Identification

Before using the quadratic formula, it's vital to rewrite your quadratic equation in the standard form \( ax^2 + bx + c = 0 \). This allows you to easily identify the corresponding coefficients \(a\), \(b\), and \(c\).
In our example, transforming \( 2x^2 = 6x + 3 \) into its standard form was necessary. We rearranged it to \( 2x^2 - 6x - 3 = 0 \). From here:

• \(a = 2\)
• \(b = -6\)
• \(c = -3\)

Recognizing these coefficients correctly ensures your application of the quadratic formula is accurate and effective. It's an essential skill for solving quadratic equations and mastering algebra.

Solving Quadratics

Once you have identified the coefficients \(a\), \(b\), and \(c\), solving the quadratic equation becomes straightforward using the quadratic formula. By substituting these values into \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), you will initiate the process of finding the solutions for \(x\).
In the given example, substituting into the formula yields:

• \( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-3)}}{2(2)} \)

The calculation proceeds logically:

• Compute the square inside the square root: \((-6)^2 = 36\)
• Calculate the expression inside the square root: \(36 + 24 = 60\)
• Plug in this value: \( x = \frac{6 \pm \sqrt{60}}{4} \)

This process reveals both possible solutions for \(x\), reflecting the intersection points of the quadratic on the x-axis. Understanding each step is crucial for solving similar problems efficiently.

Simplifying Radicals

Simplifying radicals is an important step when working with quadratic solutions, often making the final answer cleaner and more interpretable. In our example of solving the quadratic equation, we ended up with \( \sqrt{60} \).
To simplify \( \sqrt{60} \), we factor it into \( \sqrt{4 \times 15} \). Since \(4\) is a perfect square, this simplifies to:

• \( \sqrt{4 \cdot 15} = 2\sqrt{15} \)

This simplification enables us to express the solution as:

• \( x = \frac{6 \pm 2\sqrt{15}}{4} \)

Finally, further simplifying the fraction by dividing through by 2 gives:

• \( x = \frac{3 \pm \sqrt{15}}{2} \)

These simpler expressions are not only more concise but also easier to interpret and apply in subsequent mathematical or real-world problems.