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# Solving Quadratic Equations by factorization method

In this section, we are going to study three methods of solving quadratic equation, namely factorization method, completing the square method and using formula.

## Solving Quadratic Equations

We have already learnt how to solve linear equations in one, two and three variable(s). Recall that the values of the variables which satisfies a given equation are called its solution(s). In this section, we are going to study three methods of solving quadratic equation, namely factorization method, completing the square method and using formula.

## Solving a quadratic equation by factorization method.

We follow the steps provided below to solve a quadratic equation through factorization method.

Step 1 Write the equation in general form ax2 + bx +c = 0

Step 2 By splitting the middle term, factorize the given equation.

Step 3 After factorizing, the given quadratic equation can be written as product of two liner factory.

Step 4 Equate each linear factor to zero and solve for x.

These values of x gives the roots of the equation.

### Example 3.27

Solve 2x2 2√6 x + 3 = 0

### Solution

2x2 − 2√6 x + 3 = 2x2 − √6 x − √6x + 3 (by spliting the middle term)

=√2x (√2x − √3 ) − √3 (√2x − √3)= (√2x − √3)( √2x − √3)

Now, equating the factors to zero we get,

(√2x − √3)( √2x − √3)=0

√2x − √3 = 0 or √2x − √3 = 0

√2x = √3 or √2x = √3

Therefore the solution is x = √3/√2 .

### Example 3.28

Solve 2m2 + 19m + 30 = 0

### Solution

2m 2 + 19m + 30 = 2m 2 + 4m + 15m + 30 = 2m(m + 2) + 15(m + 2)

= (m + 2)(2m + 15)

Now, equating the factors to zero we get,

(m + 2)(2m + 15) = 0

m + 2 = 0  gives, m = – 2  or 2m + 15 = 0  we get, m = -15/2

Therefore the roots are -2 , -15/2

Some equations which are not quadratic can be solved by reducing them to quadratic equations by suitable substitutions. Such examples are illustrated below.

### Example 3.29

Solve x4 − 13x2 + 42 = 0

### Solution

Let x2 = a. Then, (x2)2 13x2 + 42 = a2 −13a + 42 = (a − 7)(a − 6)

Given, (a − 7)(a − 6) = 0 we get, a = 7 or 6.

Since a = x2 , x2  = 7 then, x= ± √7 or x2 = 6 we get, x = ± √6

Therefore the roots are x = ± √7 , ± √6

### Example 3.30 Therefore, the roots are x = −1 , 2.

Tags : Example, Solution | Algebra , 10th Mathematics : UNIT 3 : Algebra
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10th Mathematics : UNIT 3 : Algebra : Solving Quadratic Equations by factorization method | Example, Solution | Algebra