Last modified on August 3rd, 2023

chapter outline

 

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:

Quadratic Equation

Labels

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

Solve the quadratic equation x2 + 4x + 3 = 0 by factoring.

Solution:

Splitting the middle term, we get,
x2 + 3x + x + 3 = 0
=> x(x + 3) + 1(x + 3) = 0
=> (x +3)(x +1) = 0
=> x +3 = 0 or x + 1 = 0
x = {-3, -1}

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.

Solve the quadratic equation x2 + 8x + 7 = 0 by completing the square

Solution:

Moving the constant term to the other side of the equation by subtracting from both sides
=> x2 + 8x + 7 – 7 = 0 – 7
=> x2 + 8x = -7
Adding ${\left( \dfrac{8}{2}\right) ^{2}}$ = 16 to both sides of the equation
=> x2 + 8x +16 = -7 +16
=> x2 + 8x +16 = 9
Factor the L.H.S
=> (x + 4)2 = 9
Factoring the left side of the equation into a perfect square and square rooting both sides of the equation to obtain x
${\dfrac{8}{2}}$ = 4 and the factor is (x + 4)2
=> ${\sqrt{\left( x+4\right) ^{2}}=\sqrt{9}}$
=> x + 4 = ±3
=> x + 4 = 3 or x + 4 = -3
=> x = {-7, -1}

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

The quad equation x2 + 8x + 15 = 0 has roots α, β. Find the quadratic equation having the roots 1/α, and 1/β.

Solution:

As we know,
α + β = -b/a = -8/1 = -8 and αβ = 15/1 = 15
Now, the new equation should have its roots to be 1/α and 1/β
Their sum = 1/α + 1/β = (α + β)/αβ = -8/15
Their product = 1/αβ = 1/15
Thus, the new quadratic equation is
x2 – (1/α + 1/β)x + 1/αβ = 0
=> x2 – (-8/15) + 1/15 = 0
=> ${x^{2}-\left( \dfrac{8}{15}\right) x+\dfrac{1}{15}=0}$
Multiplying both sides by 15,
=> 15x2 + 8x + 1 = 0
Thus, the new quadratic equation is 15x2 + 8x + 1 = 0

Last modified on August 3rd, 2023

Leave a Reply

Your email address will not be published. Required fields are marked *