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SS2: MATHEMATICS - 1ST TERM

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  1. Logarithms of Numbers Less Than One
    2 Topics
  2. Theory of Logarithms
    6 Topics
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  3. Approximation & Accuracy
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  4. Sequence
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  5. Series
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  6. Geometric Progression
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  7. Quadratic Equations II
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  8. Quadratic Equations III
    1 Topic
  9. Quadratic Equations IV
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SS2: MATHEMATICS – 1ST TERM Quadratic Equations II Solve Quadratic Equations by Completing the Square
Lesson 7, Topic 1
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Solve Quadratic Equations by Completing the Square

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  • Solve Quadratic Equations by Completing the Square

It is important to note that not all quadratic expressions can be factorized and so an alternate method, Completing the Square, can be used to solve any quadratic equation of the form \( \; \scriptsize ax^2 + bx + 2 \! \! \!\)

This is referred to as the Method of Completing the Square.

Example 7.1.1:

Add the term that makes each of the following a perfect square.

i. \(\scriptsize x^2 \: – \: 20x \)           

ii.   \(\scriptsize 3x^2 \: + \: 10x \)            

iii.  \( \normalsize \frac{1}{4}\scriptsize x^2 \: – \: \normalsize \frac{2}{3}\scriptsize x \)  

iv. \(\scriptsize 4x^2 \: – \: \normalsize \frac{2}{3} \scriptsize x \) 

Solution:

i. \(\scriptsize x^2 \: – \: 20x \)   

Let \(\scriptsize x^2 – 20x + k \) be a perfect square that is equal to \(\scriptsize \left( x + a \right)^2 \) 

 

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SOLUTION (i)

⇒ \( \scriptsize x^2 \:+ \:9x \:+ \:17 = 0\)

transfer 17 to the right hand side RHS

i.e.    \( \scriptsize x^2 \:+ \:9x  = \:- 17 \)  

Complete the square for the left hand side (LHS)

⇒ Add the square of the left half of the coefficient of 9

i.e. \( \left( \frac{9}{2} \right)^2 \)

⇒ \( \scriptsize x^2 \:+ 9x \:+ \:\left( \frac{9}{2} \right)^2  = \:-17 \:+ \:\left( \frac{9}{2} \right)^2 \)  

= \( \scriptsize x^2 \:+ \:9x \:+ \:\left( \frac{81}{4} \right)  =  \frac{-17}{1} \:+ \:\left( \frac{81}{4} \right) \)  

= \( \scriptsize x^2 \:+ \:9x +\: \left( \frac{81}{4} \right) =  \frac{- 68 \:+\:81}{4} \)

= \( \scriptsize x^2\: + \:9x\: +\: \left( \frac{81}{4} \right) =  \frac{13}{4} \)

Factorise left hand side (LHS)

= \(\left(  \scriptsize x  \:+\: \frac{9}{2} \right) ^2 =  \frac{13}{4} \)

Take the square root of both sides

= \( \scriptsize x \: + \: \frac{9}{2} = \sqrt {\frac{13}{4}} \)

= \( \scriptsize x  \:+ \: \frac{9}{2} = \pm \frac{ \sqrt{13}}{2}\) 

x = \(– \frac{9}{2} \:+ \: \frac{  \sqrt{13}}{2} \: or \:  – \frac{9}{2}\: – \: \frac{  \sqrt{13}}{2}\)

= \( \frac{-9\:+ \:3.61}{2}  \: or \:  \frac{-9\:- \:3.61}{2}\)  

⇒ \( \frac{-5.39}{2}  \: or \:  \frac{-12.61}{2}\) 

⇒      x = -2.695       or  -6.305

SOLUTION (ii)

⇒ \( \scriptsize 3x^2 = \: – 9x \:+ \:6 \)

Re-arrange 

⇒ \( \scriptsize 3x^2  \: +\: 9x\:- \:6  = 0\)

Divide both sides by 3

⇒  \( \scriptsize x^2  \: + \:3x\: – \:2  = 0\)  

Transfer the constant “2” to the right hand side (RHS)

⇒  \( \scriptsize x^2  \: + \:3x = 2 \)

Complete the square for the LHS

= \( \left(  \scriptsize x  \:+\: \frac{3}{2} \right) ^2 =  \frac{2}{1} \:+ \:\frac{9}{4}\)

= \( \left(  \scriptsize x \: +\: \frac{3}{2} \right) ^2 =  \frac{8 \:+\: 9}{4} \)

= \( \left(  \scriptsize x \: +\:\frac{3}{2} \right) ^2 =  \frac{17}{4} \)

Take the square root of both sides

= \( \scriptsize x  \:+  \:\frac{3}{2} = \pm \frac{ \sqrt{17}}{2}\) 

⇒  \( \scriptsize x  =  \: – \frac{3}{2}  \pm \frac{ \sqrt{17}}{2}\) 

⇒  \( \scriptsize x  = \:  – \frac{3 \pm \sqrt{17} }{2}  \) 

x = \( \frac{-3 \:+\: \sqrt{17} }{2} \:or \:  \frac{-3 \: – \: \sqrt{17} }{2} \)

x = \(  \frac{-3 \:+\: 4.12 }{2} \:or \: \frac{- 3 \: – \: 4.12 }{2} \)

x = \( \frac{1.12 }{2} \:or \:  \frac{-7.12}{2} \)

⇒    x = 0.56       or          x = -3.56

SOLUTION (iii)

⇒ \( \scriptsize 5x^2 \:+ \:4x \: – \: 6 = 0 \)

Transfer -6 to the right hand side (RHS)

⇒     \( \scriptsize 5x^2\:+\: 4x = 6 \)   

Divide both sides by 5

⇒     \( \scriptsize x^2\: + \:\frac{4}{5}x =\frac{6}{5} \)   

Complete the square

= \( \left(  \scriptsize x  \:+ \:\frac{2}{5} \right) ^2 =  \frac{6}{5} \:+ \:\frac{4}{25}\)

= \( \left(  \scriptsize x  \:+\: \frac{2}{5} \right) ^2 =  \frac{30\: +\: 4}{25} \)

= \( \left(  \scriptsize x  \:+ \:\frac{2}{5} \right) ^2 =  \frac{34}{25} \)

Take the square root of both sides

= \( \scriptsize x  \:+ \: \frac{2}{5} = \pm \frac{ \sqrt{34}}{5}\) 

⇒ \( \scriptsize x  = \: – \frac{2}{5}  \pm \frac{ \sqrt{34}}{5}\) 

⇒  \( \scriptsize x  =   \frac{-2 \pm \sqrt{34} }{5}  \) 

x = \( \frac{-2 \:+\: \sqrt{34} }{5} \:or \:   \frac{-2 \: – \: \sqrt{34} }{5} \)

x = \(  \frac{-2 \:+\: 5.83 }{5} \:or \:  \frac{-2 \: – \: 5.83 }{5} \)

x = \( \frac{3.83 }{5} \:or \: \frac{-7.83}{5} \)

x = 0.76     or    -1.56

x = 0.8           or      -1.6

SOLUTION (iv)

⇒  \( \normalsize \frac{3}{4} \scriptsize x^2 \: – \:\normalsize \frac{2}{3} \scriptsize x \: – \: 20 = 0 \)

Clear fractions: Multiply both sides by 12

⇒  \( \scriptsize 9x^2 \:-\: 8x \:- \:240 = 0 \)  

Transfer 240 to the right hand side (RHS)

⇒  \( \scriptsize 9x^2\: – \:8x = 240  \)  

Divide both sides by 9

⇒     \( \scriptsize x^2 \: – \:  \frac{8}{9}x =\frac{240}{9} \)   

= \( \scriptsize x^2 \: – \: \frac{8}{9}x =\frac{80}{3} \)   

Complete the square

= \( \left(  \scriptsize x  \:- \:\frac{4}{9} \right) ^2 =  \frac{80}{3} \:- \:\frac{16}{81}\)

= \( \left(  \scriptsize x  \:-\: \frac{4}{9} \right) ^2 =  \frac{2160 \: – \: 16}{81} \)

= \( \left(  \scriptsize x  \:- \:\frac{4}{9} \right) ^2 =  \frac{2144}{81} \)

Take the square root of both sides

= \( \scriptsize x  \:- \: \frac{4}{9} = \pm \frac{ \sqrt{2144}}{9}\) 

⇒  \( \scriptsize x  =  \frac{4}{9} \pm \frac{46.30}{9}\) 

x = \( \frac{4}{9} \:+ \:\frac{46.30}{9} \:or \:  \frac{4}{9} \: – \: \frac{46.30}{9}\)

x = \(  \frac{4 \:+ \:46.30 }{9} \:or \:  \frac{4 \: – \:46.30}{9} \)

x = \( \frac{50.30 }{5} \:or \:  \frac{-42.30}{9} \)

x = 5.6         or      -4.7

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