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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)

$\left(s + 100 \right) \left(\frac{1500}{s} – 0.5 \right) = 1500$

$=>\left(s + 100 \right) \left(\frac{1500 – 0.5s}{s}\right) = 1500$

$=>\left(s + 100 \right) \left(1500 – 0.5s\right) = 1500s$

$=>1500s – 0.5s^{2} + 150000 – 50s = 1500s$

$=>- 0.5s^{2} + 150000 – 50s = 0$

$=> s^{2} – 300000 + 100s = 0$

$=> s^{2} + 100s – 300000 = 0$

## What is the Standard Form of a Quadratic Equation?

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 $\left(a_{1}b_{2} – a_{2}b_{1} \right) \left(b_{1}c_{2} – b_{2}c_{1} \right) = \left(a_{2}c_{1} – a_{1}c_{2} \right)^{2}$
- When $a \gt 0$, the quadratic expression $f\left(x \right) = ax^{2} + bx + c$ has a minimum value at $x = -\frac{b}{2a}
- When $a \lt 0$, the quadratic expression $f\left(x \right) = ax^{2} + bx + c$ has a maximum value at $x = -\frac{b}{2a}$
- The domain of any quadratic function is the set of all real numbers.
- For $a \gt 0$, the range of the quadratic function $f\left(x \right) = ax^{2} + bx + c$ is $[-\frac{b^{2} – 4ac}{4a}, \infty)$
- For $a \lt 0$, the range of the quadratic function $f\left(x \right) = ax^{2} + bx + c$ is : $(-\infty, -\frac{b^{2} – 4ac}{4a}]$

## Practice Problems

- Which of the following are quadratic equations?
- $\left(x – 1 \right) \left(x + 4 \right) = x^{2} + 1$
- $\left(x + 1 \right) \left(2x + 3 \right) = x^{2} + 2$
- $\left(2x + 1 \right) \left(3x – 4 \right) = 6x^{2} + 3$

- Is -2x^{2} = \left(5 – x \right) \left(2x – \frac{2}{5} \right) a quadratic equation?
- If $\frac{1}{2}$ is a root of the equation $x^{2} + kx – \frac{5}{4} = 0$, then what is the value of $k$ ?
- What constant must be added or subtracted to the quadratic equation $9x^{2} + \frac{3}{4}x – \sqrt{2} = 0$ to make it a perfect square?
- What is the discriminant of the quadratic equation $x^{2} – 4x + 3$?
- Convert the following quadratic equations to vertex form
- $x^{2} – 3x + 5$
- $2x^{2} + 7x + 9$

- Convert the following quadratic equations to intercept form
- $2x^{2} – 12x + 5$
- $-3x^{2} – 15x + 9$

## FAQs

### What is the definition of a quadratic equation?

A quadratic equation in math is a second-degree equation of the form $ax^{2} + bx + c = 0$. Here $a$, $b$, are the coefficients, $c$ is the constant term, and x is the variable. Since the variable $x$ is of the second degree, there are two roots or answers for this quadratic equation.

### What is discriminant in a quadratic equation?

The value $b^{2} – 4ac$ is called the discriminant and is designated as $\text{D}$. The discriminant is part of the quadratic formula. The discriminants help us to find the nature of the roots of the quadratic equation, without actually finding the roots of the quadratic equation.

### How are quadratic equations different from linear equations?

A linear degree is an equation of a single degree and one variable, and a quadratic equation is an equation in two degrees and a single variable. A linear equation is of the form $ax + b = 0$ and a quadratic equation is of the form $ax^{2} + bx + c = 0$. A linear equation has a single root and a quadratic equation has two roots or two answers. Also, a quadratic equation is a product of two linear equations.

## Conclusion

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 quadratic equations have two zeroes or roots which sometimes can be imaginary numbers also.

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