GED Mathematical Reasoning: Quadratic Equations II – Solve Using Quadratic Formula

Recall that a quadratic equation can be solved using factoring, but that this method doesn’t always work since not all quadratic expressions can be factored. Another method for solving a quadratic equation is to use the quadratic formula.

If given a quadratic equation in standard form ax^2 + bx + c =  0, the solutions of the equation can be found using the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Although it may look intimidating at first, the good thing about using the quadratic formula to solve a quadratic equation is that it always works.

And one more note before we continue on: The “plus or minus” symbol may be new to you. It’s simply a shorthand way of stating that we will add AND ALSO subtract two values in the numerator.

Let’s use the quadratic formula to see how it can also be used to solve this quadratic equation.

  • The first step is to put the equation in standard form.
  • The next step is to identify a, b, and c
  • Then we will substitute the values for a, b, and c into the quadratic formula to find the x-value or values that make the equation true.

 

Example 1

Solve by factoring: 2x^2 + 15x = 8

 

You may recognize the equation in the previous chapter as it was used to show the process of solving a quadratic formula by factoring. The solutions are:  x = \frac{1}{2}, -8. Let’s use the quadratic formula to see how it can also be used to solve this quadratic equation.

The first step is to put the equation in standard form. To do so, we need to subtract 8 from both sides to get all the terms on one side of the equal sign and zero on the other.

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The next step is to identify a, b, and c. a is the coefficient of the squared term, b is the coefficient of the x term, and c is the constant term. Be sure to include the sign preceding the term with the coefficient. If the sign is addition, the coefficient is positive. Is the sign is subtraction, then the coefficient is negative. algeb86

Now we will substitute the values for a, b, and c into the quadratic formula to find the x-value or values that make the equation true.

When we substitute 2 for a , 15 for b, and -8 for c we get:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \frac{-15 \pm \sqrt{{(15)}^2 - 4(2)(-8)}}{2(2)}

Step 4 is to simplify, making sure to follow the order of operations.

x = \frac{-15 \pm \sqrt{{(15)}^2 - 4(2)(-8)}}{2(2)}

x = \frac{-15 \pm \sqrt{225- 4(2)(-8)}}{2(2)}

x = \frac{-15 \pm \sqrt{225+ 64}}{2(2)}

x = \frac{-15 \pm \sqrt{289}}{2(2)}

x = \frac{-15 \pm 17}{2(2)}

Here’s our two possible solutions.

x = \frac{-15 + 17}{2(2)}, \frac{-15 - 17}{2(2)}

x = \frac{2}{4}, \frac{-32}{4}

x = \frac{1}{2}, -8

So the solutions are x = \frac{1}{2} and x = -8 , which are the two solutions we expected.

 

Example 2

Solve using the quadratic formula: x^2 + x - 5 = 0

 

We can try to solve the quadratic equation in example 2 by factoring, but we will quickly learn that the quadratic expression does not factor, making this the perfect situation for using the quadratic formula.

The first step of the process is already complete since the equation is already in standard form with all the terms on one side of the equal sign and zero on the other.

The next step is to identify a, b, and c.

a = 1
b = 1
c = -5

Now we substitute the values for a, b, and c into the quadratic formula.

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-5)}}{2(1)}

Step 4 is to simplify, making sure to follow the order of operations.

x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-5)}}{2(1)}

x = \frac{-1 \pm \sqrt{1 - 4(1)(-5)}}{2(1)}

x = \frac{-1 \pm \sqrt{1 + 20}}{2(1)}

x = \frac{-1 \pm \sqrt{21}}{2(1)}

Now for the square root step. 21 is not a perfect square, so we will leave this part as the square root of 21.

x = \frac{-1 + \sqrt{21}}{2}, \frac{-1 - \sqrt{21}}{2}

As a side note, although we can estimate the decimal version of the square root of 21 with our calculator, we typically leave the square root “as is” since this is the most accurate representation.

The two solutions are: x = \frac{-1 + \sqrt{21}}{2} and \frac{-1 - \sqrt{21}}{2}

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