What is a Quadratic Formula?

What are the applications of a Quadratic Formula?

The quadratic formula simplifies solution of quadratic equation problems.

What are quadratic equations?

If you equate a trinomial (polynomial with three terms) to zero, you get a quadratic equation. It takes the form

    \[a x^{2}+b x+c=0\]

When we are talking about the quadratic formula, we mean:

    \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\]

We can simply plug in the values of a, b, and c and use our calculators to get the two possible values of x. In this tutorial, we will develop the formula from the general quadratic equation and then use it to solve some examples.

How do I derive the quadratic formula?

We have many approaches to deriving a quadratic formula. There is the standard method, substitution, using algebraic identities, and Lagrange resolvents. In this tutorial, we will use the standard method, which makes use of completing the square technique. You can familiarize yourself with completing the square method as a prerequisite to this section.

How do I derive the quadratic formula using the standard method? 

We begin with the general quadratic equation:

    \[a x^{2}+b x+c=0\]

Taking the c term to the right of the equation gives:

    \[a x^{2}+b x=-c\]

Now, let’s divide every term by the leading coefficient a:

    \[x^{2}+\frac{b}{a} x=-\frac{s}{a}\]

Next, we aim at writing the terms on the left side of the equation in the form

    \[x^{2}+2 m x+m^{2}\]

That way, we can use algebraic identities to express the left terms in x only without powers.

To achieve that, add

    \[\left(\frac{b}{2 a}\right)^{2}\]

to both sides:

    \[x^{2}+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^{2}=-\frac{c}{a}+\left(\frac{b}{2 a}\right)^{2}\]

Using the identity

    \[(a+b)(a+b)=a^{2}+a b+b a+b^{2}=a^{2}+2 a b+b^{2} \text { gives: }\]

    \[\left(x+\frac{b}{2 a}\right)^{2}=-\frac{\varepsilon}{a}+\left(\frac{b}{2 a}\right)^{2}\]

Solve for x by taking the square root of both sides:

    \[x+\frac{b}{2 a}=\pm \sqrt{-\frac{c}{a}+\left(\frac{b}{2 a}\right)^{2}}\]

We need to have only x on the left side:

    \[x=-\frac{b}{2 a} \pm \sqrt{-\frac{c}{a}+\left(\frac{b}{2 a}\right)^{2}}\]

The equation is beginning to take shape, but we need to tidy things up a bit. By multiplying the right side by

    \[\frac{2 a}{2 a}\]

, we get:

    \[x=\frac{-b \pm \sqrt{-x(2 a)^{2}+\left(\frac{b}{\lambda_{e}}\right)^{2} \times(2 a)^{2}}}{2 a}\]

By further simplification, we get:

    \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\]

 , which is the quadratic formula.

How to use the quadratic formula? 

Now, we have the formula ready for use. How does it make our work easier in solving quadratic equations? Let’s find out using an example.

Example 1:

Solve for x in 

    \[3 x^{2}+11 x=-10\]

Solution 

We start by rearranging the equation to take the form 

    \[a x^{2}+b x+c=0\]

    \[3 x^{2}+6 x+10=0\]

 , where a=3, b=11, and c=10

Applying the formula:

    \[\begin{array}{c}x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}=\frac{-11+\sqrt{11^{2}-4 \times 3 \times 10}}{2 \times 3}=\frac{-11 \pm \sqrt{121-120}}{6}=\frac{-6 \pm \sqrt{1}}{6}=\frac{-6+1}{6} \\x=-\frac{5}{6} \text { or }-\frac{7}{6}\end{array}\]

Example 2:

Solve for x in

    \[x^{2}+3 x-4=0\]

Solution 

The equation is already in the form that we need. a=1, b=3, and c=-4

Applying the formula:

    \[\begin{array}{c}x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}=\frac{-3 \pm \sqrt{3^{2}-4 \times 1 \times-4}}{2 \times 1}=\frac{-3 \pm \sqrt{9+16}}{2}=\frac{-3 \pm \sqrt{25}}{2}=\frac{-3+5}{2} \\x=1 \text { or }-4\end{array}\]

Example 3:

Solve for x in

    \[2 x^{2}+3 x-5=0\]

Solution 

Quickly identify that a=2, b=3, and c=-5.

Applying the formula:

    \[\begin{array}{c}x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{-3 \pm \sqrt{3^{2}-4 \times 2 \times-5}}{2 \times 2}=\frac{-3 \pm \sqrt{9+40}}{4}=\frac{-3 \pm \sqrt{49}}{4}=\frac{-3 \pm 7}{4} \\x=-\frac{10}{4} \text { or } 1\end{array}\]

Remarks 

Using quadratic formula in solving quadratic equations is very easy and helps you save time. However, you have to be keen and make the right substitutions for the constants a, b, and c. If asked to find only positive values of x or leave your answer to a given number of decimal places, do just that to get all the scores.

You will use quadratic formula very often in solving quadratic equations that result from many mathematical problems. So, practice with it to make it part of you.

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