The concepts of quadratic equations are covered here. The quadratic equation PDF is also provided here which the students can download and refer offline. It should be noted this topic is one of the most important topics in maths. Among the various concepts of quadratic equation, the key points that are covered here are:

- Quadratic Equation Definition
- The standard form of Quadratic Equation
- The solution for Quadratic equation using factoring the quadratic equation and completing the square method
- Quadratic Equation Formula
- About Discriminant
- Nature of Roots of a Quadratic Equation
- The relation between Coefficient and Roots of the Quadratic Equation

## Quadratic Equation Definition:

A quadratic equation is defined as any equation which has a degree of 2. We have already discussed polynomial in Algebra. A polynomial equation having a degree of two is known as the Quadratic equation. “Quad” means four and “Quadratic” means ‘to make a square.’

## Standard Form of Quadratic equation:

The standard form of Quadratic equation is \(ax^{2}+bx+c = 0\) Where, x is an unknown and a,b,c are the constants such that \(a \neq 0\) If a becomes 0 then the equation will become a linear equation rather than a quadratic equation.

*Example:* Determine all the quadratic equation given below-

(i) \(3x^{2}+ 5x + 1 = 0\)

(ii) \( 13x + 7 = 0\)

(iii) \(3x^{5}+ 32x^{2} + 7 = 0\)

(iv) \(9x^{2} – 36 = 0\)

*Solution:* (i) and (iv) forms the quadratic equation.

## Solution for Quadratic equation:

As a quadratic equation has degree of 2, hence the number of solution that will satisfy the given equation will also be equal to 2. There can be more than one method for finding the solution for the quadratic equation:

*Sridhara’s*Formula

## Formula for Quadratic Equation:

Also known as *Sridhara’s* Method, which is most common for finding the roots of a quadratic equation \(ax^{2}+bx+c = 0\), which is given as:

\(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)

here \(\pm\) represent both the values (one positive and one negative), hence there are two possible roots for the equation.

### Additional Quadratic Related Articles:

### Discriminant (D):

The term \(b^{2}-4ac\) is known as discriminant (D), as it differentiates two roots one being positive and other being negative.

So the formula becomes: \(\frac{-b \pm \sqrt{D}}{2a}\), (where \(D = b^{2}-4ac\))

## Nature of Roots of a Quadratic Equation:

Depending upon the value of discriminant, the quadratic equation can have 3 different types of roots, namely:

S.no |
Value of Discriminant (D) |
Nature of Roots |

1 | \(b^{2}-4ac > 0\) | Two distinct Real roots |

2 | \(b^{2}-4ac = 0\) | Two equal roots |

3 | \(b^{2}-4ac < 0\) | Two distinct Complex Roots |

### Example Question:

Here a = 2, b = 7 and c = 3 Finding the value of D, \(D = b^{2}-4ac = 7^{2}- 4\times 2 \times 3 = 49 – 24\)\(\Rightarrow D = 25\) As \(D >0\), the equation has two distinct real roots. \(x = \frac{-b \pm \sqrt{D}}{2a}\) \(\Rightarrow x = \frac{-7 \pm \sqrt{25}}{2 \times 2}\) \(\Rightarrow x = \frac{-7 \pm 5}{4}\) \(\Rightarrow x = \frac{-7 – 5}{4}\; or \; \frac{-7 + 5}{4}\) \(\Rightarrow x = -3 \; or \; \frac{-1}{2}\) |

## Relation between Coefficient and Roots of the Quadratic equation:

Consider \(\large \alpha \) and \(\large \beta\) be the roots of the quadratic equation of the form \(\large ax^{2}+ bx+ c = 0 \) …………….….(i)

Then, the function can be represented as the multiple of roots, given as, \(\large (x- \alpha)(x- \beta) = 0\) ………………..(ii)

On comparing equation (i) and (ii), we have,

\(ax^{2}+ bx+ c = (x- \alpha)(x- \beta) = 0 \)

Or, \(ax^{2}+ bx+ c = x^{2} – (\alpha + \beta)x + \alpha \beta = 0 \)

Taking ‘a’ as common from the first term, we have

\(a \left [x^{2}+ \left ( \frac{b}{a} \right ) x + \left ( \frac{c}{a} \right ) \right ] = x^{2} – (\alpha + \beta)x + \alpha \beta = 0 \)

The above equation can also be written as \(\left [x^{2}+ \left ( \frac{b}{a} \right ) x + \left ( \frac{c}{a} \right ) \right ] = x^{2} – (\alpha + \beta)x + \alpha \beta\)

On comparing the terms,

\(\large \mathbf{- (\alpha + \beta) = \frac{b}{a}}\)

and, \(\large \mathbf{\alpha \beta = \frac{c}{a}}\)

Therefore the relation is given as,

Sum of the roots of the quadratic equation = \(\large \mathbf{ \frac{-b}{a}}\) |
Product of the roots of the quadratic equation = \(\frac{c}{a}\) |

### Practice Question:

** Question: **Find the maximum possible product of the roots of the quadratic equation \(x^{2} – 11x + s = 0\), where s is a natural number. (The roots of the equation belongs to integers).

** Solution: **Given \(x^{2} – 11x + s = 0\),

The general form of the quadratic equation is \(ax^{2} + bx + c = 0\),

Comparing the coefficients of the equation, we have a = 1, b = -11, and c = s,

Let \(\alpha\) & \(\beta\) be the roots of the quadratic equation.

We know that the sum of the roots of the quadratic equation is given as \(\frac{-b}{a}\), and the product of the roots is given as \(\frac{c}{a}\).

Thus, \(\alpha + \beta = -\frac{-11}{1} = 11\) ………………..(i)

\(\alpha \beta = \frac{s}{1} = s\) ………………..(ii)

Observing equation (ii),

We know s belongs to natural number, which means \(s>0\)

\(\alpha \beta = s > 0\)

This implies that, both the roots can either take both the positive values or the negative values——— (Condition 1)

Now observing equation (i),

The sum of roots (\(\alpha + \beta\))) belongs to integers, which is equal to 11.

Thus, the roots can take both the positive values OR One positive and one negative (but not both the negative values, as the sum of two negative number can never be equal to 11)—————(Condition 2)

Noticing both the conditions, we can definitely state that both the roots need to be positive.

Now the sum of two positive integers equal to 11, will give us the possible following pairs of the roots:

(1,10) , (2,9), (3,8), (4,7), (5,6) Product of the root is maximum for (5,6).

Thus, the roots of the equation, for the given condition is (5,6) or (6,5). Study more about Polynomials, Complex roots of quadratic equation etc with BYJU’S.

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