Quadratic Equations

An equation is a statement connecting two expressions by an equality sign. A polynomial is an algebraic expression whose variable powers are whole numbers. A quadratic equation is a second-order polynomial equation in one variable in the form of $a{x}^2+bx+c=0$, where $a,b$ are real number constants and $a\ne 0$.

Quadratic Equation Definition

A second-order polynomial equation in one variable in the form $a{x}^2+bx+c=0,$ where $a,b$ are real number constants and $a\ne 0$ is called a quadratic equation. The standard form of a quadratic equation is $a{x}^2+bx+c=0$
For example, $3x^2+2x+1=0, 3x^2-2x+1=0, 3x^2+1=0, 3x^2+2x=0, \sqrt{3}{x}^2=5$

The quadratic expressions can be of three types: ax^2+c,ax^2+bx,ax^2+bx+c.

Roots of a Quadratic Equation

As the degree of a quadratic equation is $2$, there will be at most $2$ roots for a quadratic equation. The roots of the equation can be found using various methods like

• Factorisation

We can factorise the polynomial $a{x}^2+bx+c$ in the form $a(x-\alpha )(x-\beta )$. $\alpha$ and $\beta$ are the roots of the equation.

• Completing the square method

An expression of the form $(x+p{)}^2$ is a perfect square. We can convert a given quadratic equation into a perfect square as below:

$a{x}^2+bx+c=0$
$\Rightarrow a\left(x^2+\frac{b}{ax}\right)+c=0$
$\Rightarrow a\left(x^2+\frac{b}{ax}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\right)+c=0$
$\Rightarrow a{\left(x+\frac{b}{2a}\right)}^2+c-\frac{b^2}{4a^2}=0$
$\Rightarrow a{\left(x+\frac{b}{2a}\right)}^2=\frac{b^2}{4a^2}-c$

• Quadratic formula

The value of $x$ can be found using the quadratic formula as $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Here, $\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}$

Graph of a Quadratic Equation

All quadratic functions can be plotted on a two-dimensional plane by finding their value for $x$.

Since quadratic is a second-order function, the shape of the graph will be a parabola.



For a quadratic equation in the form $a{x}^2+bx+c=0$, the opening of the curve will be dependent on the leading coefficient, that is, the coefficient of $x^2$.



If $a>0$, then the curve will open upwards

If $a<0$, then the curve will open downwards

Minimum and Maximum Value of a Quadratic Function

A quadratic function's maximum and minimum value will be at the vertex of the graph. The points are shown below:



If $a>0$, then the function will have a minimum point. If $a<0$, then the function will have a maximum point. The $x-$coordinate of the vertex of the graph can be found as

$h=\frac{-b}{2a}$

$k$ can be found by substituting $h$ for $x$ in $a{x}^2+bx+c$.

The maximum and minimum values of a quadratic equation can be found by completing the square method.

Case 1: $a>0$

As $a{\left(x+\frac{b}{2a}\right)}^2$is either positive or $0$, the minimum value of $a{x}^2+bx+c$ occurs when ${\left(x+\frac{b}{2a}\right)}^2$is $0$. Hence, the minimum value of the function is $c-\frac{b^2}{4a^2}$.

Case 2: $a<0$

As $a{\left(x+\frac{b}{2a}\right)}^2$ is either negative or $0$, the maximum value of $a{x}^2+bx+c$ occurs when $a{\left(x+\frac{b}{2a}\right)}^2$ is $0$. Hence, the maximum value of the function is $c-\frac{b^2}{4a^2}$.

Nature of Roots of Quadratic Equation

The nature of the roots of a quadratic equation can be identified before finding the roots. We can check that by observing the discriminant.
For a quadratic equation in standard form

The discriminant $△=b^2-4ac$

• If $△>0$, then the function has two real distinct roots

• If $△=0$, then the function has two real equal roots

• If $△<0$, then the function has two imaginary roots

• If $△$ is a perfect square, then the roots will be rational and unequal.

The points where the graph touches the $x-$axis depend on the values of $x$.
If the $x$ is real, the graph touches at most two points on $x-$axis.
If $x$ is $0$, then the graph touches the $x-$axis at only one point.
If $x$ is imaginary, then the graph will not touch $x-$axis at any points.

When $a>0$, all the graphs will open upwards for the three conditions of the discriminant.



When $a<0$, all the graphs will open upwards for the three conditions of the discriminant.




If $a>0$, the value of $a(x-\alpha )(x-\beta )$ is

• $0$, when $x=\alpha$ or $\beta$
• Negative, when x lies between α and β
• Positive otherwise

The minimum value of the expression occurs at $x=\frac{\alpha +\beta }{2}$

If $a<0$, the value of $a(x-\alpha )(x-\beta )$ is

• $0$, when $x=\alpha$ or $\beta$
• Positive, when x lies between α and β
• Negative otherwise

The maximum value of the expression occurs at $x=\frac{\alpha +\beta }{2}$


The expression $a(x-\alpha )(x-\beta )$ always have the same sign as $a$, except when $x$ lies between $\alpha$ and $\beta$, where $\alpha$ and $\beta$ area real values of intercepts on the $x-$axis. If $\alpha$ and $\beta$ are the roots of the quadratic equation, then the value of the quadratic expression changes sign only when the curve crosses the $x-$axis. As the curve crosses the $x-$axis at the point where its value becomes zero, the curve changes sign at $\alpha$, where $a{\alpha }^2+b\alpha +c=0$.



The coefficients of the quadratic equation can hint at the roots of the equation. For the equation $a{x}^2+bx+c=0$,

• If $b=0$, the roots are equal in magnitude but opposite in sign
  
  
• If $a=c$, the roots are reciprocals of each other

• If $a$ and $c$ have the same sign, but b has the opposite sign, the roots are positive

• If $a$ and $b$ have the same sign, but c has the opposite sign, the roots have opposite signs.

Sum and Products of the Roots of the Quadratic Equation

We can factorise the polynomial $a{x}^2+bx+c$ in the form $a(x-\alpha )(x-\beta )$.

Then, if $a{x}^2+bx+c=a(x-\alpha )(x-\beta )=a{x}^2-a(\alpha +\beta )+a\alpha \beta$

⇒ The sum of the roots of the quadratic equation, $\alpha +\beta =\frac{-b}{a}$
And the product of the roots of the quadratic equation $\alpha \beta =\frac{c}{a}$

Solved Examples

Q1. Find the number of real roots of the equation ${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$.

Ans: To find the nature of the roots, we need to find the discriminant of the function.
Discriminant $△=b^2-4ac$
Here, $a=1, b=1$ and $c=1$
$\Rightarrow △=1^2-4\times 1\times 1=1-4=-3$ 
Since the discriminant is less than $0$, the two roots will be imaginary; hence, there will be no real roots for the given function.

Q2. Find the roots of the equation ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathit{x}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{0}$ using the quadratic formula.

Ans: The roots of the quadratic equation can be found by the quadratic method as $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Here, $a=1, b=-5, c=5$
$\Rightarrow x=\frac{-1\pm \sqrt{(-5{)}^2-4\times 1\times 5}}{2\times 1}=\frac{-1\pm \sqrt{25-20}}{2}=\frac{-1\pm \sqrt{5}}{2}$
Hence, the roots are $\alpha =\frac{-1+\sqrt{5}}{2}$ and $\beta =\frac{-1-\sqrt{5}}{2}$

Q3. For a quadratic equation of ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{10}\mathit{x}\mathbf{+}\mathbf{25}$, will there be a maximum or minimum value? Find the value.

Ans: We can see that the leading coefficient $a>0$, and so the graph of the function will be opening upwards and hence the function will have a minimum value.
We can factorise the given quadratic equation as $x^2-10x+25=x^2-2\times 5x+5^2=(x-5{)}^2=(x-5\left)\right(x-5)$
Now the function is in the form $a(x-\alpha )(x-\beta )$
The minimum value of the expression occurs at $x=\frac{\alpha +\beta }{2}=\frac{5+5}{2}=5$
The $y$ coordinate can be found by substituting $5$ instead of $x$ in the equation, and we get $y=0$
Hence, the minimum value of the function occurs at $(5,0)$.

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