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# Quadratic Equation Definition (With Different Forms & Examples)

November 2, 2022

This post is also available in: हिन्दी (Hindi)

There are various types of equations based on their degrees, such as linear, quadratic, cubic, quartic, etc. Each of these equations has its own significance in solving real-world problems. For example, when a rocket is launched, its path is described by a quadratic equation.

Quadratic equations are the equations of degree $2$. Let’s understand the quadratic equation definition, the methods of solving it, and their use in solving problems.

## Quadratic Equation Definition

Quadratic equations are the polynomial equations of degree $2$ in one variable of the form $p\left(x \right) = ax^{2} + bx + c = 0$ where $a$, $b$, $c$ are real numbers and $a \ne 0$. It is the general form of a quadratic equation where $a$ is called the leading coefficient and $c$ is called the absolute term of $p\left(x \right)$. The values of $x$ satisfying the quadratic equation are the roots or zeroes of the quadratic equation. These two zeroes or roots of a quadratic equation are denoted by $\alpha$ and $\beta$.

## Framing a Quadratic Equation

In real math problems, the quadratic equations are presented in different forms such as

• $\left(x – 1 \right) \left(x + 2 \right) = 0$
•  $-x^{2} = -5x + 3$
• $2x \left(x + 1 \right) = 8x$
• $x^{3} = x \left(x^{2} + 2x – 1 \right)$
• $\frac {2}{x – 1} + \frac{3}{x + 2} = 1$

All of these equations need to be transformed into a standard form of the quadratic equation before performing further operations.

Let’s understand the process of framing quadratic equations from real-world problems.

Ex 1: A natural number, when increased by $12$, equals $160$ times its reciprocal. Find the number.

Let a natural number be $x$.

Then its reciprocal = $\frac{1}{x}$.

Number increased by $12$ = $x + 12$

$160$ times of reciprocal = $160 \times \frac{1}{x} = \frac{160}{x}$

Therefore,  $x + 12 = \frac{160}{x}$

$x + 12 = \frac{160}{x} => \left(x + 12 \right) \times x = 160 => x^{2} + 12x = 160 => x^{2} + 12x – 160 = 0$

Ex 2: A train, traveling at a uniform speed would have taken $48$ minutes less to travel the distance of $360 \text{km}$ if its speed were $5 \text{km/h}$ more. Find the original speed of the train.

Let the original speed of the train = $x \text{km/h}$

Distance = $360 \text{km}$

Usual time taken = $\frac{360}{x} \text{h}$ $\left(\text{Time = } \frac{\text{Distance}}{\text{Speed}} \right)$

Increased speed of train = $\left(x + 5 \right) \text{km/h}$

Time taken with increased speed = $\frac{360}{x + 5} \text{h}$

Therefore, $\frac{360}{x} – \frac{360}{x + 5} = \frac{48}{60}$

$=>360\left(\frac{1}{x} – \frac{1}{x + 5} \right) = \frac{4}{5}$

$=>360\left(\frac{x + 5 – x}{x\left(x + 5 \right)} \right) = \frac{4}{5}$

$=>\frac{5}{x\left(x + 5 \right)} = \frac{4}{5} \times \frac{1}{360}$

$=>\frac{5}{x\left(x + 5 \right)} = \frac{1}{450}$

$=>\frac{1}{x\left(x + 5 \right)} = \frac{1}{450} \times \frac{1}{5}$

$=>\frac{1}{x^{2} + 5x} = \frac{1}{2250}$

$=>x^{2} + 5x = 2250$

Ex 3: If Zeba were younger by $5$ years than she really is, then the square of her age (in years) would have been $11$ more than five times her actual age. What is her age now?

Let the present age of Zeba = $x$ years

Then age in case $5$ years younger = $x – 5$ years

$\left(x – 5 \right)^{2} = 5 \times x = x^{2} – 2 \times x \times 5 + 5^{2} = 5x$

$=>x^{2} – 10x + 25 = 5x =>x^{2} – 10x + 25 – 5x = 0 => x^{2} – 15x + 25 = 0$

Ex 4: At $t$ minutes past $2 \text{p.m.}$, the time needed by the minute’s hand of a clock to show $3 \text{p.m.}$ was found to be $3$ minutes less than $\frac{t^{2}}{4}$ minutes. Find $t$.

We know that the time between $2 \text{p.m.}$ and $3 \text{p.m}$ = 1 \text{h } = 60 \text{min}$Given that, at$t$minute past$2 \text{p.m.}$the time needed by the minute hand of a clock to show$3 \text{p.m.}$was found to be$3$min less than$\frac{t^{2}}{4}$min i.e,$t + \frac{t^{2}}{4} – 3 = 60=>4t + t^{2} – 12 = 240=>t^{2} + 4t – 252 = 0$Ex 5: A plane left$30$minutes late than its scheduled time and in order to reach the destination$1500$km away in time, it had to increase its speed by$100 \text{km/h}$from the usual speed. Find its usual speed. Distance covered by the plane =$1500 \text{km}$Let the usual speed be$s \text{km/h}$Let the usual time taken be$t \text{h}$The plane left$30$minutes late and reached on time. Therefore, time taken in this case =$\left(t – 0.5 \right) h$Speed increased by$100 \text{km/h}$Therefore, speed in this case =$\left(s + 100 \right) \text{km/h}s \times t = 1500$——–(1)$\left(s + 100 \right) \left(t – 0.5 \right) = 1500$———(2) From (1),$t = \frac{1500}{s}$Substituting the value of$t$in (2)$\left(s + 100 \right) \left(\frac{1500}{s} – 0.5 \right) = 1500=>\left(s + 100 \right) \left(\frac{1500 – 0.5s}{s}\right) = 1500=>\left(s + 100 \right) \left(1500 – 0.5s\right) = 1500s=>1500s – 0.5s^{2} + 150000 – 50s = 1500s=>- 0.5s^{2} + 150000 – 50s = 0=> s^{2} – 300000 + 100s = 0=> s^{2} + 100s – 300000 = 0$Is your child struggling with Maths? We can help! 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The standard form of a quadratic equation is$ax^{2} + bx + c = 0$, where •$a$is the coefficient of$x^{2}$•$b$is the coefficient of$x$•$c$is the constant Note: • The name Quadratic comes from “quad” meaning square because the variable gets squared ($x^{2}).
• A quadratic equation will always have two zeroes or roots.
• The nature of zeroes or roots may be either real or imaginary.

This equation is called ‘quadratic’ as its degree is $2$ because ‘quad’ means ‘square’.

Apart from the standard form of a quadratic equation, a quadratic equation can be written in other forms.

• Vertex Form: $a \left(x – h \right)^{2} + k = 0$
• Intercept Form: $a \left(x – p \right) \left(x – q \right) = 0$

Let us learn more about the standard form of a quadratic equation and let us see how to convert one form of a quadratic equation into another.

### Converting Standard Form of Quadratic Equation into Vertex Form

The standard form of a quadratic equation $ax^{2} + bx + c = 0$ can be converted into the vertex form $a \left(x – h \right)^{2} + k = 0$ (where $\left(h, k \right)$ is the vertex of the quadratic function $f\left(x \right) = a \left(x – h \right)^{2} + k$.

Note: The value of $a$ is the same in both equations.

Let $y = ax^{2} + bx + c$

$=> y – c = ax^{2} + bx$

$=> y – c = a\left(x^{2} + \frac{b}{a}x \right)$

$=> y – c = a\left(x^{2} + 2 \times x \times \frac{b}{2a} + \frac{b^{2}}{4a^{2}} – \frac{b^{2}}{4a^{2}}\right)$

$=> y – c = a\left(x^{2} + 2 \times x \times \frac{b}{2a} + \frac{b^{2}}{4a^{2}} \right) – \frac{b^{2}}{4a}$

$=> y – c = a\left(x + \frac{b}{2a} \right)^{2} – \frac{b^{2}}{4a}$

$=> y – c + \frac{b^{2}}{4a} = a\left(x + \frac{b}{2a} \right)^{2}$

$=> y – \left(c – \frac{b^{2}}{4a} \right) = a\left(x + \frac{b}{2a} \right)^{2}$

$=> y – \left(\frac{4ac – b^{2}}{4a} \right) = a\left(x – \left(-\frac{b}{2a} \right) \right)$

This is in the form $y – k = a\left(x – h \right)^{2}$, where $k = \frac{4ac – b^{2}}{4a}$ and $h = -\frac{b}{2a}$ and it is called the vertex form of the quadratic equation.

Note: $\left(h, k \right) = \left(-\frac{b}{2a}, \frac{4ac – b^{2}}{4a}\right)$ are the coordinates of the vertex of a parabola.

### Converting Standard Form of Quadratic Equation into Intercept Form

The standard form of a quadratic equation $ax^{2} + bx + c = 0$ can be converted into the intercept form $y = a\left(x – p \right) \left(x – q \right)$, where $b = -a\left(p + q \right)$ and $c = pq$.

Let $y = ax^{2} + bx + c$

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

Splitting $\frac{b}{a}$ as $-\left(p + q \right)$, such that $pq = \frac{c}{a}$

$y = a\left(x^{2} – (p + q)x + pq\right)$

$=>y = a\left(x^{2} – px – qx + pq\right)$

$=>y = a\left( x\left(x – p \right) – q\left(x – p \right)\right)$

$=>y = a \left(x – p \right) \left(x – q \right)$

## What is Discriminant in Quadratic Equation?

The discriminant is widely used in the case of quadratic equations and is used to find the nature of the roots. Though finding a discriminant for any polynomial is not so easy, there are formulas to find the discriminant of quadratic and cubic equations that make our work easier.

Discriminant of a polynomial in math is a function of the coefficients of the polynomial. It is helpful in determining what type of solutions a polynomial equation has without actually finding them. i.e., it discriminates the solutions of the equation (as equal and unequal; real and nonreal) and hence the name “discriminant”. It is usually denoted by $\triangle$ or $D$. The value of the discriminant can be any real number (i.e., either positive, negative, or 0).

For a quadratic equation $ax^{2} + bc + c = 0$, the discriminant is given by $\text{D or }\triangle = b^{2} – 4ac$.

## Formulas Related to Quadratic Equations

The following list of important formulas is helpful to solve quadratic equations.

• The quadratic equation in its standard form is $ax^{2} + bx + c = 0$
• The discriminant of the quadratic equation is $D = b^{2} – 4ac$
• For $D \gt 0$ the roots are real and distinct
• For $D = 0$ the roots are real and equal
• For $D \lt 0$ the real roots do not exist, or the roots are imaginary
• The formula to find the roots of the quadratic equation is $x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}$
• The sum of the roots of a quadratic equation is $\alpha + \beta = -\frac{b}{a}$
• The product of the roots of the quadratic equation is $\alpha \beta = \frac{c}{a}$.
• The quadratic equation whose roots are $\alpha$ and $\beta$, is $x^{2} – \left(\alpha + \beta \right)x + \alpha \beta = 0$
• The condition for the quadratic equations $a_{1}x^{2} + b_{1}x + c_{1} = 0$, and $a_{2}x^{2} + b_{2}x + c_{2} = 0$ having the same roots is $\left(a_{1}b_{2} – a_{2}b_{1} \right) \left(b_{1}c_{2} – b_{2}c_{1} \right) = \left(a_{2}c_{1} – a_{1}c_{2} \right)^{2}$