Table of Contents

Last modified on August 3rd, 2023

‘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:

**Labels**

ax^{2} + bx + c, here ‘a’ and ‘b’ are the coefficients, ‘x’ is the variable, ‘c’ is a constant

ax^{2} + 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 x^{2} will be a non-zero term (a ≠ 0)

Some examples of quadratic equations are:

x^{2} + 2x – 15 = 0, here a = 1, b = 2, and c =-15

x^{2} – 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:

2x^{2} -36 = x

Rearranging this equation, we get

2x^{2 }– x – 36 = 0, this equation is now in the standard form

Similarly, for the equation

(a – 4)^{2} – 9 = 0

=> a^{2} – 8a + 16 – 9 = 0

=> a^{2} – 8a + 7 = 0

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:

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}}

(b^{2} – 4ac) is called the discriminant because it discriminates between the possible answers:

- When b
^{2}– 4ac > 0, we get two real roots - When b
^{2}– 4ac = 0, we get one real root - When b
^{2}– 4ac < 0, we get two complex roots

For example, the two roots of the equation are:

x^{2} + 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 x^{2} + 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 **b ^{2} − 4ac**) is negative, we get a pair of Complex solutions

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

Here, the discriminant is negative

b^{2} − 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}

For a quadratic equation in standard form ax^{2} + 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

=> x^{2} + (a + b)x + ab = 0

=> x^{2} + 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 x ^{2} + 4x + 3 = 0 by factoring.**

Solution:

Splitting the middle term, we get,

x^{2} + 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}

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 ax^{2} + bx + c = 0, follow the following steps:

**Step 1**: Rewrite the equation in the form ax^{2} + 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 x^{2} must be 1. If it is not one, divide the entire equation by that number to make the coefficient of x^{2} 1.

For example, in the equation 4x^{2} – 12x – 4 = 12

First, divide the entire equation by 4

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

=> x^{2} – 3x – 1 = 3

Then follow the steps discussed above to get the result.

**Solve the quadratic equation x ^{2} + 8x + 7 = 0 by completing the square**

Solution:

Moving the constant term to the other side of the equation by subtracting from both sides

=> x^{2} + 8x + 7 – 7 = 0 – 7

=> x^{2} + 8x = -7

Adding ${\left( \dfrac{8}{2}\right) ^{2}}$ = 16 to both sides of the equation

=> x^{2} + 8x +16 = -7 +16

=> x^{2} + 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}

Quadratic equations can also be solved graphically as a function y = ax^{2} + 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.

The sum of the roots of the quadratic equation is equal to the negative of the coefficient of x divided by the coefficient of x^{2}. For the quadratic equation ax^{2} + bx + c = 0

Hence,

- α + β = -b/a = – Coefficient of x/Coefficient of x
^{2}

The product of the root of the equation is equal to the constant term divided by the coefficient of the x^{2}^{}

Hence,

- αβ = c/a = Constant term/ Coefficient of x
^{2}

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

x^{2} – (α + β)x + αβ = 0

**The quad equation x ^{2} + 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

x^{2} – (1/α + 1/β)x + 1/αβ = 0

=> x^{2} – (-8/15) + 1/15 = 0

=> ${x^{2}-\left( \dfrac{8}{15}\right) x+\dfrac{1}{15}=0}$

Multiplying both sides by 15,

=> 15x^{2} + 8x + 1 = 0

Thus, the new quadratic equation is 15x^{2} + 8x + 1 = 0

Last modified on August 3rd, 2023