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Math Example--Quadratics--Completing the Square: Example 7

Math Example | Quadratics | Completing the Square: Example 7

This is part of a collection of math examples that focus on quadratic expressions, functions, and equations.

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The next section provides background information on solving quadratic equations. This section provides other ways of solving quadratic equations, in addition to completing the square.


Solving Quadratic Equations

Quadratic equations are usually written as a quadratic expression in standard form equal to zero.

a x squared plus b x plus c equals 0

A quadratic equation can have two, one, or zero real number solutions. There are several ways to solve a quadratic. These are the methods we’ll be looking at:

  • The Quadratic Formula

  • Factoring

  • Graphing the Quadratic Function

  • Completing the Square

 Let’s look at the first method, which will work for any quadratic equation.

The Quadratic Formula

When a quadratic equation is written in standard form, like the one shown below, then you can use the quadratic formula to find the solutions to the equation.

a x squared plus b x plus c equals 0

Use the a, b, and c values from the quadratic equation and plug them into the quadratic formula:

Quadratic Formula. Used to solve a quadratic equation. The roots of the quadratic are either real or complex. The values for a, b, and c come from the quadratic function in standard form.

The quadratic formula: x equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction

To learn more about using the quadratic formula to solve quadratic equations, click on this link. This slide show includes a video overview of the quadratic formula and a number of detailed math examples.

Before using the quadratic formula, calculate the discriminant, which is the term under the square root sign of the quadratic formula.

Discriminant. The part of the quadratic formula used to determine whether a quadratic equation has real roots, and how many.

To see examples of using the discriminant, click on this link.

Factoring

You’ve already seen how to factor quadratic expressions into the product of linear terms. That same idea can be used to factor certain quadratic expressions in order to find the solutions to the equation.

A factored quadratic equation will look something like this:

open parentheses x minus a, close parentheses open parentheses x minus b close parentheses equals 0

The solutions to this equation are x = a and x = b. 

A more simplified version of a factored quadratic can look like this:

x open parentheses x minus a, close parentheses equals 0

The solutions to this equation are x = 0 and x = a. 

The previous two examples both had two solutions. There is a factored form that has one solution:

open parentheses x minus a, close parentheses squared equals 0

This is the case of the binomial squared. In this case the solution to the equation is x = a. 

The simplest example of the binomial squared is this:

x squared equals 0

The solution to this is x = 0.

If a quadratic cannot be easily factored, then you should use the quadratic formula or graph the quadratic.

To see examples of using factoring to solve a quadratic equation, click on this link.

Solving by Graphing

A visual approach to solving quadratic equations is to graph the parabola. There are three cases to look at.

Case 1: Two solutions. If the graph of the parabola intersects the x-axis twice, then there are two solutions. 

Suppose you are solving this quadratic equation:

 x squared minus 6 x plus 8 equals 0

To find the solution graphically, then graph the corresponding quadratic function.

y equals x squared minus 6 x plus 8

The graph of a parabola that intersects the x-axis at x = 2 and x = 4.

Notice that this parabola intersects the x-axis at x = 2 and x = 4. Those are the solutions to the quadratic equation. In fact, you can rewrite the quadratic in factored form:

open parentheses x minus 2 close parentheses open parentheses x minus 4 close parentheses equals 0

Case 2: One solution. If the graph of the parabola intersects the x-axis once, then there is only one real number solution. 

Suppose you are solving this quadratic equation:

 x squared plus 4 x plus 4 equals 0

To find the solution graphically, then graph the corresponding quadratic function.

y equals x squared plus 4 x plus 4

Graph of a Parabola that intersects the x-axis at x = negative 2.

Notice that this parabola intersects the x-axis at x = -2. This is the solution to the quadratic equation. In fact, you can rewrite the quadratic as a binomial squared:

open parentheses x plus 2 close parentheses squared equals 0

Case 3: No real solutions. If the graph of the parabola doesn’t intersect the x-axis, then there are no real solutions to the quadratic equation. 

Suppose you are solving this quadratic equation:

 x squared minus 4 x plus 6 equals 0

To find the solution graphically, then graph the corresponding quadratic function.

y equals x squared minus 4 x plus 6

Graph of a parabola that doesn't intersect the x-axis, indicating that it has no real roots.

Notice that this parabola doesn’t intersect the x-axis. When this happens, the quadratic equation doesn’t have real number solutions. It does, however, have complex number solutions, which you can find using the quadratic formula. 

Summary of Solving by Graphing. When a parabola intersects the x-axis, then the parabola has at least one real number solution. These intersection points are also referred to as:

  • x-intercepts

  • Zeros of the Quadratic Function

  • Roots of the Quadratic Equation

To learn more about solving a quadratic equation graphically, click on the following link. This includes a video tutorial and several worked-out math examples.

 

Solving Quadratic Equations

Example 1

The parabola with the equation shown below intersects the line with equation y = 16 at two points, A and B. What is the length of segment AB?

y equals open parentheses x minus 8 close parentheses squared

For a question of this type, where references are made to graphs and intersection points, it’s best to draw a diagram to get a better understanding of the problem. The equation shown is of a parabola in vertex form that intersects y = 16. Sketch that.

The graph of the parabola y = the quantity x minus 8 squared and the graph of y = 16. The points A and B are where the graphs intersect.

We know the y coordinates for A and B; in both cases it’s 16. To find the corresponding x-coordinates, solve this equation:

Solving a quadratic equation

The x-coordinate for point A is x = 4 and the for B it’s x = 12. So the distance from A to B is the difference, or 8.

Example 2

In the quadratic equation below, a is a nonzero constant. The vertex of the parabola has coordinates (c, d). Which of the following is equal to d?

y equals a open parentheses x minus 3 close parentheses open parentheses x plus 5 close parentheses

When a quadratic is written in standard form, the vertex has these coordinates:

Showing the relationship between the values a, b, and c in the standard form and h and k in vertex form.

Write the function in standard form. 

Writing a quadratic function in standard form.

Now find the corresponding x and y coordinates with this equation.

Finding the coordinates of the vertex of a parabola when the equation is written in standard form.

Example 3

The parabola whose equation is shown below intersects the graph of y = x at (0, 0) and (a, a). What is the value of a?

y equals 2 x squared minus 5 x

To get a better understanding of this problem, draw a diagram. 

Graph of the parabola y = 2x squared minus 5x and y = x.

To find the value of a, solve the following equation:

Solving a quadratic equation to find the value of parameter a.

Completing the Square

This way of solving a quadratic involves rewriting the quadratic equation in standard form and rewriting it in a way to find the roots using a square term. Let’s look at an example.

Quadratic equation in standard form.

This is a quadratic equation in standard form. To use the technique of completing the square, take the two x-terms and place them in parentheses.

Completing the square.

The goal is to take the expression in parentheses and write it as an expanded square term. Recall that a linear term squared has this type of expansion.

An example of binomial expansion.

So, if we want to take the previous expression and rewrite it as an expanded, it would look like this:

Completing the square

What value goes in the box to make this a squared expression? The number is 4. Do you see it?

This is what the squared term looks like.

Completing the square 

Let’s go back to the original quadratic equation.

Completing the square.

All we need to do is add a 42 to the term in parentheses. But we can’t just add this value without subtracting that same value. Otherwise, the equation would no longer be true. So this is how to write the equation.

{'mathml':'<math style=\'font-family:stix;font-size:36px;\' xmlns=\'http://www.w3.org/1998/Math/MathML\'><mstyle mathsize=\'36px\'><mfenced><mrow><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>8</mn><mi>x</mi><mo mathcolor=\'#FF0000\'>+</mo><msup><mn mathcolor=\'#FF0000\'>4</mn><mn mathcolor=\'#FF0000\'>2</mn></msup></mrow></mfenced><mo>+</mo><mn>7</mn><mo mathcolor=\'#FF0000\'>-</mo><msup><mn mathcolor=\'#FF0000\'>4</mn><mn mathcolor=\'#FF0000\'>2</mn></msup><mo>=</mo><mn>0</mn></mstyle></math>','truncated':false}

Notice that the term 42 is added to the expression in parentheses but also subtracted outside the parentheses. This keeps the original equation intact, but now you have a square term in parentheses. You can rewrite the equation this way.

{'mathml':'<math style=\'font-family:stix;font-size:36px;\' xmlns=\'http://www.w3.org/1998/Math/MathML\'><mstyle mathsize=\'36px\'><mtable columnspacing=\'0px\' columnalign=\'right center left\'><mtr><mtd><msup><mfenced><mrow><mi>x</mi><mo>+</mo><mn>4</mn></mrow></mfenced><mn>2</mn></msup><mo>+</mo><mn>7</mn><mo>-</mo><mn>16</mn></mtd><mtd><mo>=</mo></mtd><mtd><mn>0</mn></mtd></mtr><mtr><mtd><msup><mfenced><mrow><mi>x</mi><mo>+</mo><mn>4</mn></mrow></mfenced><mn>2</mn></msup><mo>-</mo><mn>9</mn></mtd><mtd><mo>=</mo></mtd><mtd><mn>0</mn></mtd></mtr><mtr><mtd/><mtd/><mtd/></mtr></mtable></mstyle></math>','truncated':false}

Now it’s in a form where you can isolate the variable to solve for x. Follow these steps.

{'mathml':'<math style=\'font-family:stix;font-size:36px;\' xmlns=\'http://www.w3.org/1998/Math/MathML\'><mstyle mathsize=\'36px\'><mtable columnspacing=\'0px\' columnalign=\'right center left\'><mtr><mtd><msup><mfenced><mrow><mi>x</mi><mo>+</mo><mn>4</mn></mrow></mfenced><mn>2</mn></msup><mo>-</mo><mn>9</mn></mtd><mtd><mo>=</mo></mtd><mtd><mn>0</mn></mtd></mtr><mtr><mtd><msup><mfenced><mrow><mi>x</mi><mo>+</mo><mn>4</mn></mrow></mfenced><mn>2</mn></msup></mtd><mtd><mo>=</mo></mtd><mtd><mn>9</mn></mtd></mtr><mtr><mtd><msqrt><msup><mfenced><mrow><mi>x</mi><mo>+</mo><mn>4</mn></mrow></mfenced><mn>2</mn></msup></msqrt></mtd><mtd><mo>=</mo></mtd><mtd><mo>±</mo><msqrt><mn>3</mn></msqrt></mtd></mtr><mtr><mtd><mi>x</mi><mo>+</mo><mn>4</mn></mtd><mtd><mo>=</mo></mtd><mtd><mo>±</mo><msqrt><mn>3</mn></msqrt></mtd></mtr><mtr><mtd><mi>x</mi></mtd><mtd><mo>=</mo></mtd><mtd><mo>-</mo><mn>4</mn><mo>±</mo><msqrt><mn>3</mn></msqrt></mtd></mtr><mtr><mtd/><mtd/><mtd/></mtr></mtable></mstyle></math>','truncated':false}

The solution is identical to what you would get had you used the quadratic formula. But with some quadratic equations, completing the square provides a useful alternative.


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Common Core Standards CCSS.MATH.CONTENT.HSA.SSE.B.3.B, CCSS.MATH.CONTENT.HSA.REI.B.4.A, CCSS.MATH.CONTENT.HSA.REI.B.4.B
Grade Range 9 - 11
Curriculum Nodes Algebra
    • Quadratic Functions and Equations
        • Quadratic Equations and Functions
Copyright Year 2021
Keywords completing the square, quadratic equations, solving quadratic equations