Quadratic Equation

‘Quad’ means square, and the variable is squared (order 2). Thus, a quadratic equation is an algebraic equation of the second degree written in the form:

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ax² + bx + c, here ‘a’ and ‘b’ are the coefficients, ‘x’ is the variable, ‘c’ is a constant

ax² + bx + c, here ‘a’ and ‘b’ are the coefficients, ‘x’ is the variable, ‘c’ is a constant

This is the standard form of the quadratic equation. For an equation to be quadratic, the coefficient of x² will be a non-zero term (a ≠ 0)

Some examples of quadratic equations are:

x² + 2x – 15 = 0, here a = 1, b = 2, and c =-15

x² – 49x = 0, here a = 1, b = -49, and c = 0

Sometimes the quadratic equations are outside the standard form and are disguised. In such cases, they were arranged and brought into the standard form. Some examples of such instances are shown below:

2x² -36 = x

Rearranging this equation, we get

2x² – x – 36 = 0, this equation is now in the standard form

Similarly, for the equation

(a – 4)² – 9 = 0

=> a² – 8a + 16 – 9 = 0

=> a² – 8a + 7 = 0

How to Solve Quadratic Equations

The solutions to the quadratic equations are its two roots, also called zeros. The simplest way to find the two roots is by using the quadratic formula:

By Quadratic Formula

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

The ‘±’ means we need to do a ‘+’ and ‘-‘operations separately to get the two solutions:

• ${x=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a} }$
• ${x=\dfrac{-b+\sqrt{b^{2}+4ac}}{2a}}

 (b² – 4ac) is called the discriminant because it discriminates between the possible answers:

• When b² – 4ac > 0, we get two real roots
• When b² – 4ac = 0, we get one real root
• When b² – 4ac < 0, we get two complex roots

For example, the two roots of the equation are:

x² + 2x – 15 = 0, here a = 1, b = 2, c = -15

= ${\dfrac{-2\pm \sqrt{4+60}}{2}}$

= ${\dfrac{-2\pm \sqrt{64}}{2}}$

= ${\dfrac{-2\pm 8}{2}}$

= ${\dfrac{-2+8}{2}}$ or ${\dfrac{-2-8}{2}}$

= {3, -5}

Thus, 3 and -5 are the two roots of the quadratic equation x² + 2x – 15 = 0 because each of them satisfies the equation:
When x = 3,

(3)² + 2(3) – 15

= 9 + 6 – 15

= 0

When x = -5

(-5)² + 2(-5) – 15

 = 25 -10 -15

 = 0

When the Discriminant is Negative

When the discriminant (the value b² − 4ac) is negative, we get a pair of Complex solutions

Let us consider the quadratic equation 5x² + 2x + 1 = 0, here a = 5, b = 2, c = 1

Here, the discriminant is negative

b² − 4ac

=> (2)² – 4 × 5 × 1

=> 4 – 20 => -16

${\sqrt{-16}}$ = 4i (here i is the imaginary number ${\sqrt{-1}}$

So,

=> ${x=\dfrac{-2+4i}{10}}$

=> x = {-0.2 + 0.4i, -0.2 – 0.4i}

By Factoring

For a quadratic equation in standard form ax² + bx + c = 0, follow the following steps:

Step 1: Split the middle term into two terms in a way such that the product of the terms is the constant term

=> x² + (a + b)x + ab = 0

=> x² + ax + bx +ab = 0

Step 2: Take common to obtain the required factors

=> x(x + a) + b(x + a) = 0

=> (x + a)(x + b) = 0

Method of Completing the Square

In this method, it is necessary to find the constant term that will enable factoring the trinomial into two identical factors. For a quadratic equation in standard form ax² + bx + c = 0, follow the following steps:

Step 1: Rewrite the equation in the form ax² + bx = c

Step 2: Add ${\left(\dfrac{b}{2}\right) ^{2}}$ to both sides of the equation

=> ${x^{2}+bx+\left( \dfrac{b}{2}\right) ^{2}=c+\left( \dfrac{b}{2}\right) ^{2}}$

Step 3: Factor the left side of the equation into a perfect square

${\left( x+\dfrac{b}{2}\right) ^{2}=c+\left( \dfrac{b}{2}\right) ^{2}}$

Step 4: Square root both sides of the equation to obtain x

${x+\dfrac{b}{2}=\pm \sqrt{c+\left( \dfrac{b}{2}\right) ^{2}}}$

To solve a quadratic equation by this method, the coefficient of x² must be 1. If it is not one, divide the entire equation by that number to make the coefficient of x² 1.

For example, in the equation 4x² – 12x – 4 = 12

First, divide the entire equation by 4

=> ${\dfrac{4x^{2}-12x-4}{4}=\dfrac{12}{4}}$

=> x² – 3x – 1 = 3

Then follow the steps discussed above to get the result.

Graphing Quadratic Equation

Quadratic equations can also be solved graphically as a function y = ax² + bx + c. By solving and then substituting the values of x in the equations, we can obtain the values of y. It will give us multiple points, which can be presented in the coordinate axis to obtain a parabola-shaped graph for the quadratic equation.

The points where the graph cuts the horizontal x-axis (x-intercepts) are the solutions of the quadratic equation. These points can also be obtained algebraically by equalizing the y value to 0 in the function and then solving for x.

Some common terms related to the graph of a quadratic function are:

• Vertex: It is the turning point of the parabola
• Axis of Symmetry: It is an imaginary line that divides the parabola into two equal halves
• Maximum: When the vertex is the highest point in the graph
• Minimum: When the vertex is the lowest point in the graph
• Domain: It is the set of all x-values that makes the function true
• Range: It is the set of all y-values obtained by substituting the values of x
• Axis of Symmetry: It is an imaginary line that divides the parabola into two equal halves

To learn in detail how to graph a quadratic function, click here.

Roots of Quadratic Equation

The sum of the roots of the quadratic equation is equal to the negative of the coefficient of x divided by the coefficient of x². For the quadratic equation ax² + bx + c = 0

Hence,

• α + β = -b/a = – Coefficient of x/Coefficient of x²

The product of the root of the equation is equal to the constant term divided by the coefficient of the x²

Hence,

• αβ = c/a = Constant term/ Coefficient of x²

If α, β, are the roots of the quadratic equation, then the quadratic equation can be written as:

x² – (α + β)x + αβ = 0