Problem 62 - Algebra 1

Question

Solve the quadratic equation by finding square roots or by using the quadratic formula. Explain why you chose the method. $$6 x^{2}+20 x+5=0$$

Step-by-Step Solution

Answer

The solutions to the quadratic equation \(6x^{2}+20x+5=0\) are \(x_{1,2}= -\frac{10 \pm \sqrt{70}}{6}\)

Step 1: Write down the quadratic equation

The given quadratic equation is \(6x^{2}+20x+5=0\)

Step 2: Identify a, b, c from the equation

From the given equation, we can identify the values of \(a\), \(b\), and \(c\) for the quadratic formula \(-\frac{b \pm \sqrt{b^{2}-4ac}}{2a}\) as \(a=6\), \(b=20\), and \(c=5\)

Step 3: Apply the quadratic formula

Substitute the values \(a=6\), \(b=20\), and \(c=5\) into the quadratic formula \(-\frac{b \pm \sqrt{b^{2}-4ac}}{2a}\) to get the solutions

Step 4: Compute the values under square root

Calculate the value under the square root \(b^{2}-4ac = (20)^{2}-4*6*5 = 400 -120 = 280\)

Step 5: Apply the quadratic formula

Apply the quadratic formula with the values \(b=20\), \(a=6\), and \(\sqrt{280}\) to get the solutions. So, \(x_{1,2}= -\frac{20 \pm \sqrt{280}}{2*6}\)

Step 6: Simplify

On simplifying the above equation, we get \(x_{1,2}= -\frac{10 \pm \sqrt{70}}{6}\)

Key Concepts

Quadratic Formula

The quadratic formula is a useful tool to solve quadratic equations of the form \(ax^2 + bx + c = 0\). This formula is versatile and can solve any quadratic equation, whether it can be factored or not. The formula is expressed as:

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

This formula provides two potential solutions due to the \(\pm\) symbol, representing the two possible roots of the equation. To use the quadratic formula, identify the coefficients \(a\), \(b\), and \(c\) from the equation.
In the equation \(6x^2 + 20x + 5 = 0\), we have \(a = 6\), \(b = 20\), and \(c = 5\). Substituting these values into the formula reveals the solutions for \(x\). Calculating \(b^2 - 4ac\) helps determine if the roots are real or complex. Here, the value is 280, implying real and distinct roots.

Solving Equations

Solving quadratic equations involves finding the values of \(x\) that make the equation true. There are several techniques for doing this:

• Factoring, if the equation is factorable
• Completing the square
• Using the quadratic formula
• Graphing

In the case of \(6x^2 + 20x + 5 = 0\), we chose the quadratic formula because it's straightforward and guaranteed to work when \(b^2 - 4ac\) is positive or zero.
First, substitute \(a\), \(b\), and \(c\) into the formula. Then, solve for \(x\) using the computed square root from \(b^2 - 4ac = 280\). Divide by \(2a\) to obtain the solutions \(x_1\) and \(x_2\). The solutions offer a way to understand the behavior of the quadratic equation graphically and algebraically.

Square Roots

Square roots are an essential part of solving quadratic equations, especially when using the quadratic formula. The square root symbol \(\sqrt{}\) represents a number that, when multiplied by itself, yields the original number under the root sign.
In the process of solving \(6x^2 + 20x + 5 = 0\), we encountered \(\sqrt{280}\). This value arises from the discriminant, \(b^2 - 4ac\), which determines the nature of the roots:

• If \(b^2 - 4ac > 0\), the equation has two distinct real roots.
• If \(b^2 - 4ac = 0\), the equation has exactly one real root.
• If \(b^2 - 4ac < 0\), the equation has complex roots.

For \(\sqrt{280}\), simplifying the expression results in smaller, more manageable values, aiding in finding the exact or approximate root values. Understanding square roots is critical for grasping the process of the quadratic formula and for analyzing the solutions' properties.