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) $9x^{2} – 36 = 0$ Solution: (i) and (iv) forms the 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

Also known as Sridhara’s Method, which is mostly common for finding the roots of 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.

## Discriminant (D):

The term $b^{2}-4ac$ is known as discriminant (D), as it differentiate 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: Find the roots of the equation $2x^{2} + 7x + 3 = 0$ Solution: Given: $2x^{2} + 7x + 3 = 0$ 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 $D >0$, 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 = $\frac{c}{a}$

 Example: 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).

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