**Solving quadratic equations :**

Let f(x) = ax² + bx +c.

Clearly f(x) is a quadratic function.

The zeros of a quadratic function are nothing but the two values of "x" when f(x) = 0 or ax² + bx +c = 0.

Here,

**"ax² + bx +c = 0" is called as quadratic equation**

Finding the two zeros of a quadratic function or solving the quadratic equation are the same thing.

In other words, the zeros of a quadratic equation are the x-coordinates of the points where the parabola (graph of quadratic a function) cuts x-axis.

There are three methods to find the two zeros of a quadratic function.

They are,

1. Factoring

2. Quadratic formula

3. Completing square

let us discuss the above three methods in detail.

Here we are going to see how to solve a quadratic equation using factoring.

Generally we have two types of equations

**Factoring quadratics with a leading coefficient of 1**

In a quadratic "Leading coefficient" means "coefficient of x²".

(i) If the coefficient is 1 we have to take the constant term and we have to split it as two parts.

(ii) The product of two parts must be equal to the constant term and the simplified value must be equal to the middle term (or) x term.

(iii) Now we have to write these numbers in the form of (x + a) and (x +b)

(iv) By equating the factors equal to zero.We can find the value of x. solving quadratic equations by factoring

**Example 1:**

Solve x² + 17 x + 60 = 0

**Solution:**

**In the first step we are going to check whether we have 1 as the coefficient of x² or not. **

**Since it is 1. We are going to take the last number. That is 60 and we are going to factors of 60.**

**All terms are having positive sign. So we have to put positive sign for both factors.**

Here,

10 x 6 = 60 but 10 + 6 = 16 not 17

15 x 4 = 60 but 15 + 4 = 19 not 17

12 x 5 = 60 and 12 + 5 = 17

2 x 30 = 60 but 2 + 30 = 32 not 17

**(x + 12) (x + 5)** are the factors

x + 12 = 0 x + 5 = 0

x = -12 x = -5

Solution is {-12,-5}

This is just example of solving quadratic equations by factoring.If you need more example problems of solving quadratic equations by factoring please click the below link.

Factoring quadratic equations when a isn't 1

(i) If it is not 1 then we have to multiply the coefficient of x² by the constant term and we have to split it as two parts.

(ii) The product of two parts must be equal to the constant term and the simplified value must be equal to the middle term (or) x term.

(iii) Divide the factors by the coefficient of x². Simplify the factors by the coefficient of x² as much as possible.

(iv) Write the remaining number along with x.

**Example 2:**

Factor 2 x² + x - 6

**Solution:**

To factor this quadratic equation we have to multiply the coefficient of x² by the constant term

So that we get -12, now we have to split -12 as the multiple of two numbers.

Since the last term is having negative sign.So we have to put negative sign for the least number.

Now we have to divide the two numbers 4 and -3 by the coefficient of x² that is 2. If it is possible we can simplify otherwise we have to write the numbers along with x.

(x + 2) (2x - 3) = 0

x + 2 = 0 2 x - 3 = 0

x = -2 2 x = 3

x = 3/2

This is just one example problem to show solving quadratic equations by factoring. If you want more example please click the below link.

Please click the following links to look at the remaining two methods in detail.

You can visit the following of our pages to know more about quadratic equations.

1. Sum and product of the roots of a quadratic equation

2. Nature of the roots of a quadratic equation

After having gone through the stuff given above, we hope that the students would have understood "Solving quadratic equations".

Apart from the stuff given above, if you want to know more about "Solving quadratic equations", please click here

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