The polynomial which can be expressed in the form of ax We can find the solution of the quadratic equations by Factorization, by completing the squares and making them the perfect squares and it is also done even by the quadratic formula. Once we learn to use the formula of the quadratic roots and to find the value of the determinants, and the nature of roots can also be known. Now let us see that α and β are the roots of the quadratic equation, then it means that if we put the value of α or β in the given equation, then it satisfies the given equation. D = b ^{2} – 4 * a * c is the formula which helps to analyze the types of roots of the equation. If D = 0, then roots are real and equal, if D> 0, then roots are unequal and real, if D< 0, then roots are imaginary. |

^{2}+ bx + c = 0 , here 'x' is a variable and a, b, c are constants they are also called as quadratic coefficient, and highest power is two, in this equation there is a condition that value of 'a' is not equals to zero, because if it is equal to zero then this equation becomes linear equation. Quadratic equation has two solutions, to solve any quadratic equation we use a Quadratic Formula.

Suppose we have an equation ax^{2} + bx + c = 0, so first of all we will find roots of this equation, so by using quadratic formula,

x = -b ∓ (√(b^{2} – 4ac))/2a , are two solutions of this quadratic equation.

x = -b + (√(b^{2} – 4ac))/2a , x = -b -(√(b^{2} – 4ac))/2a .

Here (b^{2} – 4ac) is called as discriminant, it is represented by Δ.

**Case 1:** If Δ > 0 , (discriminant is positive) then in this case there are two roots and both are Real Numbers.

Roots = (-b +(√ Δ ))/2a , (-b - (√Δ))/2a .

**Case 2:** If Δ = 0, in this case we have only one real root (double root).

Root = - b / 2a .

**Case 3 **: If Δ < 0 (discriminant is negative), in this case there are two roots, they both are non real.

Root = -b / 2a + ί( √-Δ ))/2a , -b / 2a - ί(√-Δ))/2a.

^{2}+ qx + r = 0, here 'x' is a variable whise value is unknown and p, q, and r are constants where 'p' can not be equals to 0. That means p ≠ 0. Constants p, q, and r are called as quadratic coefficient, linear coefficient and constant term respectively. Here we will study

**quadratic equations by factoring**.

^{2}+ qx + r = 0, if 'c' is the root of given quadratic equation px

^{2}+ qx + r = 0.

^{2}+ qx + r = p (x - [(-q + √q

^{2}+ 4 p r) / 2p]) (x - [(-q + √q

^{2}– 4 p r) / 2q]).

^{2}+ 5x + 6 = 0,

^{2}+ 3x + 2x + 6 = 0,

^{2}+ 5x + 6 = 0 are,

^{2}+ bx + c = 0 and Quadratic Formula is given by:

⇨

**x = - b + √ (b**, its alternate form also given by:

^{2}– 4ac) / 2a⇨ x

**= 2c / -b +√ (b**.

^{2}– 4ac)Now we will see how to find factors of quadratic equation using

**Quadratic Formula**. To find factors we need to follow some steps:

Step 1: To find factor first we take a quadratic equation. Let quadratic equation is 5p

^{2}– p + 4 = 0.

Step 2: Use above formula to find factors of expression. We know that quadratic formula is given by: ⇨ x = - b + √ (b

^{2}– 4ac) / 2a. In equation the value of ‘a’ is 5, value of ‘b’ is -1 and value of ‘c’ is 4.

Step 3: Put these values in formula. On putting these values we get:

⇨ x = - b + √ (b

^{2}– 4ac) / 2a,

⇨ x = - (-1) + √ ((-1)

^{2}– 4 * 5 * 4) / 2 (5);

⇨ x = -1 + √ (-1 – 80) / 10;

⇨ x = -1 + √ 81 / 10; so here we will get two factors of given equation.

X = -1 + √ 81 / 10 and X = -1 - √ 81 / 10.

This is how we find factors of quadratic equation using quadratic formula.

**Square root property can be explained as a Square root which is the mathematical reverse of a squared exponent.**It also says that squaring of a positive or a negative number will be equals to a positive number. Square root property states that if a² = b then a = √b or a = -√b, which can also be written as, a = ±√b. Here we will have two values possibly. Both values will be said real if 'c' is positive otherwise they will be imaginary if 'c' is negative.

**. This property has to be applied to both sides of equation to keep things even-handed. If any other arithmetic operations are needed to be performed, go for them to find answer from there. Note that at last you will be getting two solutions for 'a'.**

We can solve a Quadratic Equation using square root property where we can take square root of a squared expression to eliminate exponent in the expression

We can solve a Quadratic Equation using square root property where we can take square root of a squared expression to eliminate exponent in the expression

Square root property can be used to solve equations like (x - 10)² = 4 sort. Begin solving it by taking square root on both sides to eliminate exponent part: we get x - 10 = √4. Place plus or minus symbol in front of square root of "4" to symbolize that two solutions are possible and so two values of x: x - 10 = ± √4. Perform the operation, retaining the plus or minus symbol: x - 10 = ± 2.

Now we are left with two equations given as x – 10 = 2 and x – 10 = - 2. Solve them for 'x'. We get, x = 12 or x = 8. This way we can solve even more complex equations using square root property.

We are very much aware of formula for Square of addition or subtraction of two Numbers which can be given as:

(a + b)

^{2}= a

^{2}+ b

^{2}+ 2ab and

(a - b)

^{2}= a

^{2}+ b

^{2}- 2ab,

When you have a Quadratic Equation of form ax² + bx + c which is not possible to be factorized, you can use technique called

**completing the square**. To complete the square means creating a polynomial with three terms that results into a perfect square.

To start with, rewrite the quadratic expression given as ax² + bx + c and move the constant term 'c' to right side of equation to get the form ax² + bx = - c. Divide this complete equation constant factor “a” if a≠ 1 to get x² + (b / a) x = -c / a.

Divide coefficient of 'x' i.e. (b / a) by 2 and it now becomes (b / 2a) and then square it to get (b / 2a) ². Add (b / 2a)² to both sides of equation to get:

x² + (b / a) x + (b / 2a)² = -c / a + (b / 2a) ².

Next step would we writing left side of equation as perfect square:

[x + (b / 2a)] ² = -c / a + (b / 2a) ².

For

**example**, let’s take an expression: 4x² + 16x - 20. Where, a = 4, b = 16 and c = -20.

Moving constant 'c' to right side we get 4x² + 16x = 20. Next divide both sides of equation by 4 to get: x² + 4x = 20 / 4. Taking half of 4 which is coefficient of 'x' and then squaring it to get:

(4 / 2) ² = 4,

Add 4 to your equation:

x² + 4x + 4 = 5 + 4,

or x² + 4x + 4 = 9,

Making left side a perfect square we get:

(x + 2)

^{2}= 9,

or x = 1, -5.