Question

Find the real solutions, if any, of each equation. Use any method. $$ 2+z=6 z^{2} $$

Short answer

The real solutions are \(z = \frac{2}{3}\) and \(z = -\frac{1}{2}\).

Step-by-step solution

Rewrite the Equation

First, rewrite the given equation in standard quadratic form: \[2 + z = 6z^2\]

Move All Terms to One Side

Subtract 2 and z from both sides to set the equation to zero: \[0 = 6z^2 - z - 2\] or equivalently, \[6z^2 - z - 2 = 0\]

Identify Coefficients

Identify the coefficients in the quadratic equation: \[a = 6,\; b = -1,\; c = -2\]

Apply the Quadratic Formula

Apply the quadratic formula \[z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] with the identified coefficients: \[z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4\cdot6\cdot(-2)}}{2\cdot6}\]

Simplify the Radicand

Simplify the expression under the square root (the radicand): \[(-1)^2 - 4\cdot6\cdot(-2) = 1 + 48 = 49\]

Calculate the Square Root

Evaluate the square root of 49: \[\sqrt{49} = 7\]

Solve for z

Substitute the simplified values back into the quadratic formula and solve for \(z\): \[z = \frac{1 \pm 7}{12}\] This gives two solutions: \[z = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3}\] and \[z = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2}\]

Key concepts

Solving Quadratic Equations

To solve a quadratic equation, you need to find the values of the variable that make the equation true. Quadratic equations typically have the form \[ax^2 + bx + c = 0\] where \(a, b,\) and \(c\) are constants. You can use several methods to solve these equations, including:

• Factoring
• Completing the square
• Using the Quadratic Formula
• Graphing

For the exercise \(2 + z = 6z^2\), we rearranged it to the standard form \(6z^2 - z - 2 = 0\) so we could apply the Quadratic Formula. Solving quadratics often means transforming them into this familiar form first to make further calculations manageable.

Quadratic Formula

The Quadratic Formula provides a straightforward way to find solutions to any quadratic equation, regardless of whether it can be factored. The formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here’s how it works:

• Identify the coefficients \(a, b,\) and \(c\).
• Substitute these coefficients into the formula.
• Calculate the discriminant, \(b^2 - 4ac\).
• Solve the resulting expression to find the values of \(x\).

In our specific example, we used the coefficients \(a = 6\), \(b = -1\), \(c = -2\) and found the solutions by plugging them into the formula. This is particularly useful when quadratics do not factor neatly.

Identifying Coefficients

Identifying the coefficients in a quadratic equation is a crucial step. These constants \(a, b,\) and \(c\) will be substituted into the Quadratic Formula or used in other solution methods. For the equation \(6z^2 - z - 2 = 0\), the coefficients were:

• \(a = 6\): The coefficient of \(z^2\)
• \(b = -1\): The coefficient of \(z\)
• \(c = -2\): The constant term

Accurately identifying these values ensures that the Quadratic Formula is used correctly. Any mistake here can lead to incorrect solutions, so double-check your coefficients!

Simplifying Radicals

Simplifying radicals is an essential skill when solving quadratic equations using the Quadratic Formula. In the context of our exercise, the discriminant \(b^2 - 4ac\) turned out to be 49. Simplifying this radical involved finding \(\sqrt{49} = 7\). Here are some tips to simplify radicals:

• Look for perfect square factors.
• Break down the number under the radical into its prime factors.
• Simplify the expression by taking the square root of any perfect squares.

Returning to our example, simplifying \(\sqrt{49}\) resulted in a clean integer (7), which made it easy to proceed to the next steps in solving the quadratic equation. Simplifying radicals can sometimes get tricky, especially with larger or non-perfect square numbers, but practice makes perfect!