Quadratic Formula

• Introduction
• What is a Quadratic Formula?
• Derivation of the Quadratic Formula
• Discriminant and Nature of Roots
• How to Solve Using Quadratic Formula
• Examples of Quadratic Formula
• Practice Problems
• Frequently Asked Questions

Introduction

The quadratic formula is a valuable tool for solving quadratic equations, providing a method that replaces the traditional factorization approach. Essentially, it is a formula used to determine the roots or solutions of a quadratic equation. When a quadratic equation lacks real roots, the quadratic formula comes to the rescue by facilitating the identification of imaginary roots!!

What is a Quadratic Formula?

To find the solution(s) of a quadratic equation, you can use the quadratic formula:

Here, `a`, `b`, and `c` are the coefficients from the quadratic equation  \( \color{#38761D}ax^2 + \color{#6F2DBD}bx + \color{#fb8500}c = 0 \). The symbol \(\pm\) represents both the positive and negative roots, providing two possible solutions. The discriminant, \(b^2 - 4ac\), plays a crucial role in determining the nature of the roots.

Derivation of the Quadratic Formula

The quadratic formula is derived from completing the square method. “Completing the square” is another method used for solving quadratic equations. Let's consider a quadratic equation in standard form:

\( \color{#38761D}ax^2 + \color{#6F2DBD}bx + \color{#fb8500}c = 0 \)

The goal is to rewrite this equation in the form \((x - p)^2 = q\), where \(p\) and \(q\) are constants. The steps for the derivation are as follows:

`1`. Move the constant term to the other side:

\( ax^2 + bx = -c \)

`2`. Divide both sides by \(a\):

\( x^2 + \frac{b}{a}x = -\frac{c}{a} \)

`3`. Add \(\left(\frac{b}{2a}\right)^2\) to both sides to complete the square:

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

Simplify the right side:

\( x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \)

`4`. Factorize the left side:

   \( \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \)

`5`. Take the square root of both sides:

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

`6`. Isolate \(x\):

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

This derived expression is the quadratic formula, providing the solutions for any quadratic equation.

Discriminant and Nature of Roots

The expression inside the square root (\(\color{#6F2DBD}b^2 - 4\color{#38761d}a\color{#fb8500}c\)) in the quadratic formula is called the discriminant. In maths, we can use uppercase delta (\(\Delta\)) as a symbol to write discriminant.The discriminant in a quadratic formula decides the number of roots a quadratic equation can have and the nature of its roots. The value of this discriminant can be either positive, negative or zero.

• `2` Real Roots \((b^2 - 4ac > 0)\): When the value inside the square root is positive, it implies that the quadratic graph intersects the `x`-axis at two distinct points, resulting in two real and distinct roots.

Example:

Quadratic Equation: \(x^2 - 6x + 5 = 0\)

Discriminant: \(b^2 - 4ac = (-6)^2 - 4(1)(5) = 36 - 20 = 16\)

Since the discriminant is greater than zero (\(16 > 0\)), the equation has two real and distinct roots.

• `1` Real Root \((b^2 - 4ac = 0)\): A discriminant of zero means that the square root portion of the quadratic formula becomes zero, resulting in only one real root. This situation corresponds to the parabola touching the `x`-axis at a single point.

Example:

Quadratic Equation: \(4x^2 - 4x + 1 = 0\)

Discriminant: \(b^2 - 4ac = (-4)^2 - 4(4)(1) = 0\)

Since the discriminant is zero, the equation has one real root.

• `2` Complex Roots (\(b^2 - 4ac < 0\)): In this scenario, the square root expression involves the imaginary unit \(i\), indicating that the roots are not real but rather complex conjugates of each other. The quadratic graph does not intersect the `x`-axis.

Example:

Quadratic Equation: \(2x^2 + 4x + 5 = 0\)

Discriminant: \(b^2 - 4ac = (4)^2 - 4(2)(5) = -24\)

Since the discriminant is negative (\(-24 < 0\)), the equation has two complex roots.

How to Solve Using Quadratic Formula

Let's go through an example step by step to illustrate how to solve a quadratic equation using the quadratic formula.

Example:

Solve the quadratic equation \(2x^2 - 5x + 3 = 0\) using the quadratic formula.

Step `1`: Identify the coefficients
\( a = 2, \quad b = -5, \quad c = 3 \)

Step `2`: Write down the quadratic formula
\( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \)

Step `3`: Plug in the coefficients
\( x = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4(2)(3)}}}}{{2(2)}} \)

Step `4`: Calculate the discriminant
\( \Delta = b^2 - 4ac \)
\( \Delta = (-5)^2 - 4(2)(3) \)
\( \Delta = 25 - 24 \)
\( \Delta = 1 \)

Step `5`: Determine the nature of roots
Since \(\Delta > 0\), there are two real and distinct roots.

Step `6`: Calculate the roots
\( x = \frac{{-(-5) \pm \sqrt{1}}}{{4}} \)

Step `7`: Simplify the roots
\( x = \frac{{5 \pm 1}}{{4}} \)

Step `8`: Express the solutions
The solutions are:
\( x = \frac{{5 + 1}}{{4}} = \frac{6}{4} = \frac{3}{2} \)
\( x = \frac{{5 - 1}}{{4}} = \frac{4}{4} = 1 \)
Therefore, the solutions to the quadratic equation \(2x^2 - 5x + 3 = 0\) are \(x = \frac{3}{2}\) and \(x = 1\).

Examples of Quadratic Formula

Example `1`: Solve the quadratic equation \(2x^2 + 10x + 11 = 0\) using the quadratic formula.

Solution:

To find the roots, we will use the quadratic formula:

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

For the given equation \(2x^2 + 10x + 11 = 0\), the coefficients are \(a = 2\), \(b = 10\), and \(c = 11\).

Substitute these values into the quadratic formula:

\( x = \frac{{-(10) \pm \sqrt{{(10)^2 - 4(2)(11)}}}}{{2(2)}} \)

Simplify the expression under the square root:

\( x = \frac{{-10 \pm \sqrt{{100 - 88}}}}{{4}} \)

\( x = \frac{{-10 \pm \sqrt{12}}}{{4}} \)

This leads to two solutions:

`1`. When using the positive square root:

\( x = \frac{{-10 + 2\sqrt{3}}}{{4}} =\frac{{-5 + \sqrt{3}}}{{2}}\)

`2`. When using the negative square root:

\( x = \frac{{-10 - 2\sqrt{3}}}{{4}} =\frac{{-5 - \sqrt{3}}}{{2}}\)

Therefore, the roots of the quadratic equation \(2x^2 + 10x + 11 = 0\) are \(\frac{{-5 + \sqrt{3}}}{{2}}\) and \(\frac{{-5 - \sqrt{3}}}{{2}}\).

Example `2`: Determine the roots of the quadratic equation \(3x^2 - 6x + 3 = 0\).

Solution:

To find the roots, we can use the quadratic formula:

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

For the given equation \(3x^2 - 6x + 3 = 0\), the coefficients are \(a = 3\), \(b = -6\), and \(c = 3\).

Substitute these values into the quadratic formula:

\( x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4(3)(3)}}}}{{2(3)}} \)

Simplify the expression under the square root:

\( x = \frac{{6 \pm \sqrt{{36 - 36}}}}{{6}} \)

\( x = \frac{{6 \pm \sqrt{0}}}{{6}} \)

Since the discriminant (\(b^2 - 4ac\)) is zero, there is only one solution:

\( x = \frac{6}{6} = 1 \)

Therefore, in the quadratic equation \(3x^2 - 6x + 3 = 0\), both roots are the same, and \(x = 1\).

Example `3`: Find the roots of the quadratic equation \(x^2 + 4x + 5 = 0\).

Solution:

To find the roots, we can use the quadratic formula:

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

For the given equation \(x^2 + 4x + 5 = 0\), the coefficients are \(a = 1\), \(b = 4\), and \(c = 5\).

Substitute these values into the quadratic formula:

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

Simplify the expression under the square root:

\( x = \frac{{-4 \pm \sqrt{{16 - 20}}}}{{2}} \)

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

Here, since the discriminant (\(b^2 - 4ac\)) is negative, the square root involves the imaginary unit \(i\). Simplify further:

\( x = \frac{{-4 \pm 2i}}{{2}} \)

Divide both terms by `2`:

\( x = -2 \pm i \)

Therefore, the roots of the quadratic equation \(x^2 + 4x + 5 = 0\) are complex, and they are \(x = -2 + i\) and \(x = -2 - i\).

Practice Problems

Q`1`. Solve the quadratic equation \(2x^2 - 7x + 3 = 0\). What are the roots?

a) \(x = 3, x = \frac{1}{2}\)  

b) \(x = 2, x = -\frac{3}{2}\)  

c) \(x = \frac{3}{2}, x = -2\)  

d) \(x = -1, x = \frac{3}{2}\)

Answer: a

Q`2`. Determine the roots of the quadratic equation \(4x^2 + 8x + 4 = 0\). What are the solutions?

a) \(x = 1\)  

b) \(x = -2\)  

c) \(x = -1\)  

d) \(x = -\frac{1}{2}\)

Answer: c

Q`3`. Find the roots of the quadratic equation \(x^2 + 6x + 13 = 0\).

a) \(x = -3 + 2i, x = -3 - 2i\)  

b) \(x = 1 + 3i, x = 1 - 3i\)  

c) \(x = -2 + 4i, x = -2 - 4i\)  

d) \(x = 2 - i, x = 2 + i\)

Answer: a

Q`4`. Solve the quadratic equation \(2x^2 + 4x + 4 = 0\). What are the roots?

a) \(x = -2 + i, x = -2 - i\)  

b) \(x = 1 + 2i, x = 1 - 2i\)  

c) \(x = -1 - i, x = -1 + i\)  

d) \(x = 3 - 2i, x = 3 + 2i\)

Answer: c

Q`5`. Determine the roots of the quadratic equation \(3x^2 - 10x + 7 = 0\).

a) \(x = 1, x = 7\)  

b) \(x = 2, x = 5\)  

c) \(x = \frac{1}{3}, x = 7\)  

d) \(x = \frac{7}{3}, x = 1\)

Answer: d

Frequently Asked Questions

Q`1`. Are there alternative methods for solving quadratic equations aside from the quadratic formula?

Answer: Yes, apart from the quadratic formula, there are various other methods of solving quadratic equations. The most common ones are listed below:

• Taking the square root method
• Factoring method
• Completing the square method

It is advisable to use the above methods if they suit the given quadratic equation. However, the quadratic formula is the most robust method, and the equation that can’t be solved by other methods can be solved using the quadratic formula.

Q`2`. What is the discriminant, and how does it determine the nature of the roots?

Answer: The discriminant helps us decide the number of solutions and the nature of the roots a quadratic equation can have when solved.

Q`3`. Can a quadratic equation have only one real root?

Answer: No, a quadratic equation can also have `2` real roots or `2` complex roots. If the value of the discriminant is positive, the equation will render `2` real roots. If the value of the discriminant is negative, the equation will render `2` complex roots.

Q`4`. How is the quadratic formula derived?

Answer: The quadratic formula is derived from another method of solving quadratics called the “Completing the Square” method.