Problem 34 - Algebra for College Students

Question

Use the method of completing the square to solve each quadratic equation. $$2 t^{2}-4 t+1=0$$

Step-by-Step Solution

Answer

The solutions are \( t = 1 + \frac{\sqrt{2}}{2} \) and \( t = 1 - \frac{\sqrt{2}}{2} \).

Step 1: Write the Equation in Standard Form

Ensure the quadratic equation is in standard form, which is already the case. The given quadratic equation is \(2t^2 - 4t + 1 = 0\).

Step 2: Move the Constant Term to the Other Side

Rearrange the equation so that the constant term is on the right side: \(2t^2 - 4t = -1\).

Step 3: Divide by the Coefficient of the Quadratic Term

Divide the entire equation by 2 (the coefficient of \(t^2\)) to simplify the left side: \(t^2 - 2t = -\frac{1}{2}\).

Step 4: Complete the Square

To complete the square, take half of the coefficient of \(t\), square it, and add to both sides. Half of \(-2\) is \(-1\), and squaring it gives \(1\). Add 1 to both sides: \(t^2 - 2t + 1 = -\frac{1}{2} + 1\).

Step 5: Simplify the Right Side

Combine the constants on the right: \(t^2 - 2t + 1 = \frac{1}{2}\).

Step 6: Rewrite as a Perfect Square

The left side \(t^2 - 2t + 1\) can now be written as \((t - 1)^2\). Therefore, \((t - 1)^2 = \frac{1}{2}\).

Step 7: Solve for t by Taking Square Roots

Take the square root of both sides: \(t - 1 = \pm\sqrt{\frac{1}{2}}\).

Step 8: Isolate t

Solve for \(t\) by adding 1 to both sides: \(t = 1 \pm \sqrt{\frac{1}{2}}\).

Step 9: Simplify the Expression

Recognize that \(\sqrt{\frac{1}{2}}\) is the same as \(\frac{\sqrt{2}}{2}\). Thus, \(t = 1 \pm \frac{\sqrt{2}}{2}\).

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation that involves terms up to the second degree. It usually looks like this: \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are coefficients, with \( a eq 0 \). Quadratic equations are fundamental in algebra because they govern a wide range of phenomena, from the shape of a parabola to equations encountered in physics and engineering. Whenever you see an equation involving \( x^2 \), you are dealing with a quadratic equation.

Standard Form

The standard form of a quadratic equation is crucial for applying various solution methods, such as factoring, using the quadratic formula, or completing the square. The standard form is expressed as \( ax^2 + bx + c = 0 \). If an equation isn't already in this form, you'll need to rearrange it to get there. In our example, the equation \( 2t^2 - 4t + 1 = 0 \) is already in standard form, making the subsequent steps more straightforward. Getting equations into this form helps in identifying the coefficients, which are necessary for further calculations.

Perfect Square Trinomial

A perfect square trinomial is an expression that can be written as the square of a binomial. For instance, \( (x - 1)^2 = x^2 - 2x + 1 \) is a perfect square trinomial. In completing the square, we aim to transform part of our quadratic equation into a perfect square trinomial. This makes solving the equation much easier. For example, with our quadratic \( t^2 - 2t \), we complete the square by deciding that half of \(-2\) and then squaring it gives us \(1\). Adding \(1\) to both sides allows us to rewrite the left side as \( (t - 1)^2 \), a perfect square trinomial.

Solving Quadratics

Solving quadratic equations can be done through various methods such as factoring, using the quadratic formula, or completing the square. Completing the square, as demonstrated in the solution, involves turning the quadratic part into a perfect square trinomial, solving for the variable using square roots, and simplifying the result. Here’s a quick run-through:

• First, rearrange your equation to isolate the quadratic and linear terms.
• Make sure the quadratic term has a coefficient of 1 by dividing the entire equation.
• Add the necessary value to complete the square, forming a perfect square trinomial.
• Express this trinomial as a square of a binomial.
• Finally, solve for the variable by taking square roots and isolating it.

This method is particularly useful when equations do not easily factor and when you're aiming for an exact solution. In our example, completing the square leads us to the solutions \( t = 1 \pm \frac{\sqrt{2}}{2} \). This process not only solves the equation but also reinforces understanding of algebraic structures.