Roots And The Discriminant Of A Quadratic Equation

The "Quadratic" word in quadratic equation comes from quadratic polynomial, which is a polynomial with the highest degree of 2. The term "Equation" is an expression which has a sign of equality ("=") between its left and right hand sides. So, it is written with a quadratic polynomial on LHS and 0 on the RHS of the equation.

An example of a quadratic equation is 3x² + 6x + 9 = 0. This example has 3x² + 6x + 9 on LHS, which is a quadratic polynomial and 0 on RHS. This chapter discusses definition, standard form, roots and methods to solve a quadratic equation.

Quadratic equation

A quadratic equation is an expression in which there exists a quadratic polynomial on LHS with the highest degree of 2 and 0 on RHS or in other words, if a quadratic polynomial equates to 0, then it becomes a quadratic equation.

A quadratic equation in variable x is written in the form of ax² + bx + c = 0, where a ≠ 0 and a, b, c are real numbers.

Standard form

An equation whose terms are arranged in descending order of their degrees is known as the standard form of an equation. The standard form of the quadratic equation is ax² + bx + c = 0 where a, b and c are real numbers and a ≠ 0. The constants a, b and c are called as coefficients of the equation.

More specifically, a is called a quadratic coefficient, b is a linear coefficient and c is constant coefficient.

Roots

In a quadratic equation, the value of LHS becomes equal to RHS when it is 0. LHS can be set to 0 for some values of x. When such numbers are put in the place of x to make the value of LHS 0, these specific values of x are called as roots of the quadratic equation.

Such values of x are said to be satisfying the quadratic equation or "the values of x satisfy the quadratic equation".

Methods to solve

Solving the quadratic equation means finding the roots of the quadratic equation. As explained above, roots are the values of x which satisfy the equation ax² + bx + c = 0. There are three ways to find the roots or to solve the quadratic equation.

1. Factorization
2. Completing the square
3. Quadratic formula

Factorization method

In this method, the middle term of the equation ax² + bx + c = 0 splits into two terms in such a way that the sum of the two terms equals the coefficient b and the product of the two terms equals to the coefficient c.

Here, the middle term is bx and b is the coefficient of bx. So, split the coefficient b into two terms p and q such that p + q = b and p × q = c, where p ≠ 0 and q ≠ 0.

Therefore, after splitting b the equation ax² + bx + c = 0 can be written as:

ax² + px + qx + c = 0

Now, the LHS of ax² + px + qx + c = 0 can be expressed as a product of two linear factors. Each of the two linear factors can be equated to 0 separately.

After solving each linear factor separately, we get one value of x from each of the two linear factors. These two values of x will be the roots of the quadratic equation.

Completing the square method

Following are the steps to solve the quadratic equation ax² + bx + c = 0, a ≠ 0 using completing the square method.

Step 1: Make the coefficient of x² equal to 1 by dividing the whole equation by 'a'

$\frac{\mathrm{ax²}}{\mathrm{a}}$ + $\frac{\mathrm{bx}}{\mathrm{a}}$ + $\frac{\mathrm{c}}{\mathrm{a}}$ = 0

or x² + $\frac{\mathrm{bx}}{\mathrm{a}}$ + $\frac{\mathrm{c}}{\mathrm{a}}$ = 0

Step 2: Move the constant term $\frac{\mathrm{c}}{\mathrm{a}}$ on the other side.

x² + $\frac{\mathrm{bx}}{\mathrm{a}}$ = 0 - $\frac{\mathrm{c}}{\mathrm{a}}$

x² + $\frac{\mathrm{bx}}{\mathrm{a}}$ = - $\frac{\mathrm{c}}{\mathrm{a}}$

Step 3: Half the coefficient of x i.e. $\frac{\mathrm{b}}{\mathrm{a}}$

$\frac{1}{2}\text{(}\frac{\mathrm{b}}{\mathrm{a}}\text{)}$ = $\frac{\mathrm{b}}{\mathrm{2a}}$

Step 4: Add the square of $\frac{\mathrm{b}}{\mathrm{2a}}$ on both sides of the equation

x² + $\frac{\mathrm{bx}}{\mathrm{a}}$ + ${\text{(}\frac{\mathrm{b}}{\mathrm{2a}}\text{)}}^2$ = $-\frac{\mathrm{c}}{\mathrm{a}}$ + ${\text{(}\frac{\mathrm{b}}{\mathrm{2a}}\text{)}}^2$

Step 5: Write LHS as perfect square of binomial expression and simplify RHS also.

${\text{(}\text{x}+\frac{\mathrm{b}}{\mathrm{2a}}\text{)}}^2$ = $\frac{{\text{b}}^2-\text{4ac}}{4{\text{a}}^2}$

Step 6: Take square root on both sides

$\sqrt{{\text{(}\text{x}+\frac{\mathrm{b}}{\mathrm{2a}}\text{)}}^2}$ = $\sqrt{\frac{{\text{b}}^2-\text{4ac}}{4{\text{a}}^2}}$

Step 7: On LHS, square root cancels out with the square. On RHS add ±

$\text{x}+\frac{\mathrm{b}}{\mathrm{2a}}$ = $\text{±}\sqrt{\frac{{\text{b}}^2-\text{4ac}}{4{\text{a}}^2}}$

Step 8: Shift the constant term $\frac{\mathrm{b}}{\mathrm{2a}}$ from LHS to RHS

x = $\text{±}\sqrt{\frac{{\text{b}}^2-\text{4ac}}{4{\text{a}}^2}}$ - $\frac{\mathrm{b}}{\mathrm{2a}}$

x = $\frac{\text{±}\sqrt{{\text{b}}^2-\text{4ac}}}{\mathrm{2a}}$ - $\frac{\mathrm{b}}{\mathrm{2a}}$

x = $\frac{-\text{b}\text{±}\sqrt{{\text{b}}^2-\text{4ac}}}{\mathrm{2a}}$

± in front of $\text{±}\sqrt{{\text{b}}^2-\text{4ac}}$ indicates two values with + and - separately i.e. $\sqrt{{\text{b}}^2-\text{4ac}}$ and $-\sqrt{{\text{b}}^2-\text{4ac}}$

So x has two values which can be written as $\frac{-\text{b}+\sqrt{{\text{b}}^2-\text{4ac}}}{\mathrm{2a}}$ and $\frac{-\text{b}-\sqrt{{\text{b}}^2-\text{4ac}}}{\mathrm{2a}}$

These two values of x are called as two roots of the quadratic equation.

Roots of ax² + bx + c = 0, a ≠ 0 are $\frac{-\text{b}+\sqrt{{\text{b}}^2-\text{4ac}}}{\mathrm{2a}}$ and $\frac{-\text{b}-\sqrt{{\text{b}}^2-\text{4ac}}}{\mathrm{2a}}$

Quadratic formula method

The quadratic formula method is another way to solve a quadratic equation. It is found easy to use as compared to the factorization method and completing the square method.

The roots of quadratic equation a² + bx + c = 0 are calculated using these two formulas

$\frac{-\text{b}+\sqrt{\text{D}}}{\mathrm{2a}}$ and $\frac{-\text{b}-\sqrt{\text{D}}}{\mathrm{2a}}$

Here, D is called discriminant and value of D = ${\text{b}}^2-\text{4ac}$

The values of a, b and c are filled up in the above formulas to find roots. The values obtained after solving it, will be the two roots of the quadratic equation.

Let's see next, how the quadratic formula method helps to solve a quadratic equation.