Solving the quadratic equation by completing the square method

Contents:

This article covers one of the methods used to solve the quadratic functions which is called ” completing the square method”.

The following are the key concepts that this article covers in detail.

  • What is a perfects square quadratic?
  • General forms of a perfect square quadratic.
  • How completing the square method is related to perfect square quadratic?
  • What is the number to be added and subtracted to complete the square?
  • Worked out examples solving a quadratic using “completing the square method”
  • Quiz Time.

1)Perfect square Quadratic.

A quadratic polynomial, which can be written as the product of two identical binomials, is called as a perfect square quadratic. This plays a key role in solving a quadratic equation using completing the square method.

i)General forms of perfect square quadratic equations.

There are two general form of representing a quadratic equation as a perfect square                                                                                   i)  y= x^{2}+2ax+a^{2}=(x+a)(x+a) = (x+a)^{2}.

 ii) y= x^{2}-2ax+a^{2} =(x-a)(x-a) = (x-a)^{2}.

2)Solving the quadratic using “completing the square” method.

The phrase “Completing the square” conveys that the given quadratic equation has to be transformed into a “perfect square quadratic”.

The aim is to represent any arbitrary quadratic equation in the form of a perfect square quadratic.

In order to represent it, there is a need to introduce a constant number which helps us to write any quadratic as perfect square quadratic.

ii)What is that number to be introduced to complete the square?

Let us consider the perfect square quadratic  y= x^{2}+2ax+a^{2}.We have the first terms as  x^{2} , middle term as  2\times a\times x and the last term as a^{2} .

All we need is to re-write the given quadratic similar to this standard form ensuring that the leading coefficient of the Quadratic, ie the coefficient of   x^{2} , is 1. We compare this given quadratic with the standard form  y= ax^{2}+bx +c and get the coefficients a,b, c.

The constant number that is essential to complete the square is given by the formula  \left( \dfrac {b}{2}\right) ^{2}.

We add and subtract this term  \left( \dfrac {b}{a}\right) ^{2} to the given quadratic and group the first and middle term with the  \left( \dfrac {b}{2a}\right) ^{2} to complete the square.

3)Worked out examples

Let us start with a simple example.

i) y=x^{2}-6x -16

Strategy Step-by-Step:-

Step1: First of all, replace y with 0.Identify the leading coefficient of the quadratic equation that is a.

  • In the current problem, replacing y with 0, we have  0=x^{2}-6x -16 .On one to one comparison with the standard form the quadratic equation, y= ax^{2}+bx +c we can say that the leading coefficient is a=1. .

 

Step2: Divide the equation throughout by the leading coefficient “a” to make the coefficient of   x^{2} as 1.

  • We already have the coefficient of the quadratic as 1 therefore, therefore no need to divide the entire equation with the leading coefficient.

 

Step3: Find the value of coefficient “b” of new equation .Find the value of  \left( \dfrac {b}{2}\right) ^{2} thereby.

  • On comparision with the standard form  y= ax^{2}+bx +c ,the value of b = -\dfrac {6}{1} =-6 .Therefore the value of  \left( \dfrac {b}{2}\right)   = \dfrac {-6}{2}=-3 and  value of   \left( \dfrac {b}{2}\right) ^{2} = 9.

Step4: Add and subtract the term  \left( \dfrac {b}{2}\right) ^{2} term.

  • x^{2}-6x+9-9 -16=0

 

Step5: Group the First term, Middle term and + \left( \dfrac {b}{2}\right) ^{2} term to complete the square and take the rest of the terms to the  right side of “=” sign.

  • Groupingthe  first, middle and + \left( \dfrac {b}{2}\right) ^{2}   terms as x^{2}-6x +9= 9+16

 

Step6: Write the Completed square as \left( x+\dfrac {b}{2}\right) ^{2} and solve for the variable “x” taking square root both sides.

  • x^{2}-6x +9= 9+16    can be written as \left( x-3\right) ^{2}= 25  .Hence,taking square root both sides we get \sqrt {\left( x-3\right) ^{2}}=\sqrt {25} .Thererfore \left( x-3\right) =\pm 5 and the value of  x would be3\pm 5.Hence the roots of the quadratic would be x= 8,-2.

 

Let us try a complex example.

ii) y=4x^{2}-6x -10

Strategy Step-by-Step:-

Step1: First of all, replace y with 0.Identify the leading coefficient of the quadratic equation that is a.

  • In the current problem, replacing y with 0, we have  0=4x^{2}-6x -10 .On one to one comparison with the standard form the quadratic equation, y= ax^{2}+bx +c we can say that the leading coefficient is a=4.

 

Step2: Divide the equation throughout by the leading coefficient “a” to make the coefficient of   x^{2} as 1.

  • The leading coefficient “a” of the given quadratic is 4.So we divide the equation with 4 both sides.

 

 \dfrac {4x^{2}-6x -10}{4}=\dfrac {0}{4}\\ \\ \Rightarrow x^{2}-\dfrac {6}{4}x -\dfrac {10}{4}=0\\ \\ \\ \Rightarrow x^{2}-\dfrac {3}{2}x-\dfrac {5}{2}=0

 

Step3: Find the value of coefficient “b” of new equation .Find the value of  \left( \dfrac {b}{2}\right) ^{2} thereby.

  • On comparision with the standard form  y= ax^{2}+bx +c ,the value of b = -\dfrac {3}{2} .Therefore the value of  \left( \dfrac {b}{2}\right)   = \dfrac {-\dfrac {3}{2}}{2}=-\dfrac {3}{4} and  value of   \left( \dfrac {b}{2}\right) ^{2} = \left( -\dfrac {3}{4}\right) ^{2} =\dfrac {9}{16} .

 

Step4: Add and subtract the term  \left( \dfrac {b}{2}\right) ^{2} term.

  • x^{2}-\dfrac {3}{2}x-\dfrac {5}{2}+\dfrac {9}{16}  -\dfrac {9}{16}=0

 

Step5: Group the First term, Middle term and + \left( \dfrac {b}{2}\right) ^{2} term to complete the square and take the rest of the terms to the  right side of “=” sign.

  • Groupingthe  first, middle and + \left( \dfrac {b}{2}\right) ^{2}   terms as x^{2}-\dfrac {3}{2}x +\dfrac {9}{16} = \dfrac {9}{16}+\dfrac {5}{2}= \dfrac {9}{16}+\dfrac {40}{16} = \dfrac {49}{16} (taking the common denomiantor as 16).

 

Step6: Write the Completed square as \left( x+\dfrac {b}{2}\right) ^{2} and solve for the variable “x” taking square root both sides.

  • x^{2}-\dfrac {3}{2}x +\dfrac {9}{16}= \dfrac {49}{16} can we writen as \left( x-\dfrac {3}{4}\right) ^{2}=\dfrac {49}{16}. Hence,taking square root both sides we get \sqrt{\left( x-\dfrac {3}{4}\right) ^{2}}=\sqrt{\dfrac {49}{16}}

Therefore \left( x-\dfrac {3}{4}\right) =\pm\dfrac {7}{4}.

\left x= \dfrac {3}{4}\right\pm\dfrac {7}{4}

\left x= \dfrac {3}{4}\right\ +\dfrac {7}{4} and \left x= \dfrac {3}{4}\right\ -\dfrac {7}{4}

Therefore x= \dfrac {10}{4} , -\dfrac {4}{4}

FInally,the simplified solutions would be x= \dfrac {5}{2} , -1

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