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

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.

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

**Step 1 **Write the equation in
general form *ax*^{2} + *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.

Solve** **2*x*^{2}** **−** **2√6** ***x*** **+** **3** **=** **0

2x^{2} − 2√6 x + 3 = 2x^{2}
− √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
.

Solve** **2*m*^{2}** **+** **19*m*** **+** **30** **=** **0

2*m* ^{2} +
19*m* + 30 = 2*m*^{ 2} + 4*m* + 15*m* + 30 = 2*m*(*m*
+ 2) + 15(*m* + 2)

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

Now, equating the
factors to zero we get,

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

*m *+* *2* *=* *0*
*gives,* m *= – 2* *or* *2*m *+* *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.

Solve *x*^{4}
− 13*x*^{2} + 42 = 0

Let *x*^{2}*
*=* a*. Then, (*x*^{2})^{2}* *−* *13*x*^{2}*
*+* *42 = *a*^{2} −13*a* + 42 = (*a* − 7)(*a*
− 6)

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

Since *a *=* x*^{2}*
*, *x*^{2}* *=* *7 then, *x*= ± √7 or *x*^{2}
= 6 we get, *x* = ± √6

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

Therefore, the roots are
*x *= −1* *, 2.

Tags : Example, Solution | Algebra Example, Solution | Algebra

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