Problem 61 - College Algebra

Question

Solve each equation in Exercises \(47-64\) by completing the square. $$4 x^{2}-4 x-1=0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x =\frac{1}{2} + \sqrt{\frac{1}{2}}\) and \(x =\frac{1}{2} - \sqrt{\frac{1}{2}}\)

Step 1: Divide each term by the coefficient of \(x^{2}\)

To start, we will first make the coefficient of \(x^{2}\) equal to 1, by dividing each term in the equation by 4. This yields \((x^{2} - x - \frac{1}{4})=0\)

Step 2: Rearrange the equation

Next, we rearrange the terms and leave space to add a term that will make the left side of the equation a perfect square trinomial. The equation becomes \((x^{2} - x + ?) = \frac{1}{4} + ?\) . We need to find the value that will make \(x^{2} - x + ?\) a perfect square.

Step 3: Complete the square

We complete the square on the left side of the equation by adding \(\left(\frac{-1}{2}\right)^2 =\frac{1}{4}\) to both sides. This gives us \((x^{2} - x +\frac{1}{4}) = \frac{1}{4} +\frac{1}{4}\) which simplifies to \((x - \frac{1}{2})^{2} = \frac{1}{2}\)

Step 4: Solve for \(x\)

Finally, apply the square root property, remembering to consider both the positive and negative square roots, then solve for \(x\). This yields \(x -\frac{1}{2} = \pm \sqrt{\frac{1}{2}}\) . Solving for \(x\), we get get \(x =\frac{1}{2} \pm \sqrt{\frac{1}{2}}\)

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\). They play a significant role in algebra. The main components are:\

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• \(ax^2\) is the quadratic term, where \(a\) is the coefficient, and it should never be zero.
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• \(bx\) is the linear term, with \(b\) being the coefficient.
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• \(c\) is the constant term.
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\Quadratic equations can be solved using different methods such as factoring, completing the square, or the quadratic formula. Each method has its own strategy based on the equation structure. Completing the square is often used when the quadratic equation cannot be easily factored. This technique creates a perfect square trinomial on one side of the equation, simplifying it into something more manageable to solve.

Perfect Square Trinomial

A perfect square trinomial is an expression that can be factored into a square of a binomial. It follows the structure \((x + d)^2 = x^2 + 2dx + d^2\). To form a perfect square trinomial from a quadratic expression like \(x^2 + bx\), you'll need to add \((\frac{b}{2})^2\) to make it complete. Here's how you can do it:\

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• Take half of the linear coefficient \(b\); this is \(\frac{b}{2}\).
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• Square this result to find \((\frac{b}{2})^2\).
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• Add this squared value to the original expression, transforming it into a perfect square trinomial \(x^2 + bx + (\frac{b}{2})^2\).
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\This perfect square trinomial can now be expressed as \((x + \frac{b}{2} )^2\). This conversion is a key step in completing the square method, helping simplify solving quadratic equations.

Square Root Property

The square root property is a helpful tool for solving equations formatted as \((x - p)^2 = q\). It allows us to quickly find the values of \(x\), by taking the square root of both sides:\

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• Start by isolating the square term. If you have \((x - p)^2 = q\), simply ensure that \(q\) is on one side of the equation.
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• Take the square root of both sides, keeping in mind that you will get both a positive and negative result: \(x - p = \pm \sqrt{q}\).
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• Solve for \(x\) by adding \(p\) to both sides, which gives you \(x = p \pm \sqrt{q}\).
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\In the context of completing the square, this property becomes a powerful way to find solutions once you have transformed the equation to a squared form. Always remember to consider both the positive and negative square roots, as both are valid and can lead to two different solutions for \(x\).