We have already discussed about Polynomial in Algebra. A polynomial equation having degree of two is known as Quadratic equation. “Quad” means four but there Quadratic means ‘to make 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 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) \(9x^{2} – 36 = 0\)
(i) and (iv) forms the quadratic equation. |

## Solution for Quadratic equation:

As 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 quadratic equation:

*Sridhara’s*Formula

## Formula for Quadratic Equation:

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

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

here \(\pm\)

## Discriminant (D):

The term \(b^{2}-4ac\)

So the formula becomes:

\(\frac{-b \pm \sqrt{D}}{2a}\)

## 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: Find the roots of the equation \(2x^{2} + 7x + 3 = 0\)
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 \(D >0\) \(x = \frac{-b \pm \sqrt{D}}{2a}\) \(\Rightarrow x = \frac{-7 \pm \sqrt{25}}{2 \times 2}\) \(\Rightarrow x = \frac{-7 \pm 5}{4}\) \(\Rightarrow x = \frac{-7 – 5}{4}\; or \; \frac{-7 + 5}{4}\) \(\Rightarrow x = -3 \; or \; \frac{-1}{2}\) |

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

Consider \(\large \alpha \)

Then, the function can be represented as the multiple of roots, given as,

\(\large (x- \alpha)(x- \beta) = 0\)

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 *= \(\frac{c}{a}\)

Example: Find the maximum possible product of the roots of the quadratic equation \(x^{2} – 11x + s = 0\)
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 \(\alpha\) We know that the sum of the roots of the quadratic equation is given as \(\frac{-b}{a}\) Thus, \(\alpha + \beta = -\frac{-11}{1} = 11\) \(\alpha \beta = \frac{s}{1} = s\) 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 (\(\alpha + \beta\) 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). |

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