Solving Quadratic Equations by Quadratic Formula

In the previous post, we have learned how to solve quadratic equations by factoring. In this post, we are going to learn how to solve quadratic equations using the quadratic formula. In doing this, we must identify the values of a, b, and c, in ax^2 + bx + c = 0 and substitute their values to the quadratic formula

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

Note that the value of a is the number in the term containing x^2, b is the number in the term containing x, and c is the value of the constant (without x or x^2).

The results in this calculation which are the values of x are the roots of the quadratic equation. Before you calculate using this formula, it is important that you master properties of radical numbers and how to calculate using them.

Example 1: Find the roots of x^2 - 4x - 4 = 0

Solution

From the equation, we can identify a = 1, b = -4, and c = -4.

Substituting these values in the quadratic formula, we have

x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-4)}}{2(1)}
x = \dfrac{4 \pm \sqrt{16 + 16}}{2}
x = \dfrac{4 \pm \sqrt{32}}{2}.

We know, that \sqrt{32} = \sqrt{(16)(2)} = \sqrt{16} \sqrt{2} = 4 \sqrt{2}. So, we have

x = \dfrac{4 \pm 4 \sqrt{2}}{2} = \dfrac{2(2 + \sqrt{2}}{2} = 2 \pm 2 \sqrt{2}

Therefore, we have two roots

2 + 2 \sqrt{2} or 2 - 2 \sqrt{2}

Example 2: Find the roots of 2x^2 - 6x = 15

Solution

Recall, that it easier to identify the values of a, b, and c if the quadratic equation is in the general form which is ax^2 + bx + c = 0. In order to make the right hand side of the equation above equal to 0, subtract 15 from both sides of the equation by 15. This results to

2x^2 - 6x - 15.

As we can see, a = 2, b = -6 and c =-15.

Substituting these values to the quadratic formula, we have

x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-15)}}{2(2)}

x = \dfrac{6 \pm \sqrt{36 + 120}}{4}

x = \dfrac{6 \pm \sqrt{156}}{4}.

But \sqrt{156} = \sqrt{(4)(39)} = \sqrt{4} \sqrt{39} = 2 \sqrt{39}.

Therefore,

x = \dfrac{6 \pm 2 \sqrt{39}}{4}.

Factoring out 2, we have

\dfrac{2(3 \pm \sqrt{39}}{4} = \dfrac{3 \pm \sqrt{39}}{2}

Therefore, we have two roots:

\dfrac{3 + \sqrt{39}}{2} or \dfrac{3 - \sqrt{39}}{2}

That’s it. In the next post, we are going to learn how to use quadratic equations on how to solve word problems.

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