Problem 33 - Introductory Algebra for College Students

Question

Solve quadratic equation by completing the square. \(4 x^{2}-2 x-3=0\)

Step-by-Step Solution

Answer

The solutions to the equation \(4 x^{2}-2 x-3=0\) are \(x = 1.15\) and \(x = -0.65\).

Step 1: Balancing the Equation

To make the quadratic equation easier to handle, divide the entire equation by 4 to balance it. This results in: \(x^{2}-0.5x -0.75 = 0\)

Step 2: Rearranging the equation

Rearrange the equation to express it in the form of a perfect square. A general quadratic equation can be expressed as: \((x-h)^{2}=k\). Comparing this with : \(x^{2}-0.5x + a = 0.75 + a\), we are choosing \(a\) such that \((x-0.25)^{2}\) will become possible since \((x-h)^{2} = x^{2}-2hx+h^{2}\). This gives \(a= (0.5/2)^{2}=0.0625\). Substituting \(a\) into the equation we get : \(x^{2}-0.5x + 0.0625 = 0.8125\)

Step 3: Applying the Square root property

The rearranged equation can now be solved by applying the square root property. Simplifying the equation, we get \((x-0.25)^{2}= 0.8125\). By applying square root property, this gives two solutions : \(x-0.25 =\sqrt{0.8125}\) and \(x-0.25 =-\sqrt{0.8125}\)

Step 4: Solving for x

Finally, solve for the variable \(x\) from both of these results separately. First solution is given by : \(x=\sqrt{0.8125} + 0.25 = 1.15\). Second solution is given by \(x= -\sqrt{0.8125} + 0.25 = -0.65\)

Key Concepts

Quadratic Equation

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The main feature of a quadratic equation is the squared term \(x^2\), which makes the equation parabolic when graphed. This type of equation can have one, two, or no real solutions, depending on the discriminant \(b^2 - 4ac\).
Quadratic equations are commonly solved using various methods such as factoring, using the quadratic formula, or completing the square, as demonstrated in the original exercise.
In completing the square, our aim is to rearrange the equation into a form that is easier to solve, often transforming it into a perfect square trinomial on one side of the equation.
In our original exercise, we began by simplifying the equation to \(x^2 - 0.5x - 0.75 = 0\) by dividing by 4. This helps in visualizing the steps needed to form a perfect square, which is the next core concept we'll discuss.

Perfect Square

A perfect square trinomial is an expression in the form \((x - h)^2\), where \(h\) is a constant. This transforms a quadratic expression into a simpler form, which can be more straightforward to solve.
In our example, the equation \(x^2 - 0.5x\) was manipulated by adding and subtracting the perfect square value \(0.0625\).
Here are the main steps involved in creating a perfect square:

• Identify the coefficient of \(x\), in this case, \(-0.5\).
• Divide this coefficient by 2, resulting in \(-0.25\).
• Square the result to get the perfect square component, \(0.0625\).

By including this perfect square in the expression, it transforms into \((x - 0.25)^2\), enabling us to use the square root property to solve the equation, our next concept focus.

Square Root Property

The square root property provides a direct method for solving equations once they are expressed as a perfect square on one side. If you have \((x - h)^2 = k\), taking the square root of both sides helps to isolate \(x\).
Applying this to our transformed equation \((x - 0.25)^2 = 0.8125\), we proceed as follows:

• Take the square root of both sides: \(x - 0.25 = \pm \sqrt{0.8125}\).
• This results in two potential expressions because the square root can be positive or negative.
• Add \(0.25\) to both sides to solve for \(x\).

Consequently, you obtain:\[x = 0.25 + \sqrt{0.8125} = 1.15\] and \[x = 0.25 - \sqrt{0.8125} = -0.65\]. Thus, the solutions to the quadratic equation are \(x = 1.15\) and \(x = -0.65\), demonstrating the effectiveness of the square root property in finding solutions with ease.