Topic Content:
- 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 5.1.1:
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 \(\scriptsize \left( x + a \right)^2 \)
⇒ \(\scriptsize x^2 – 20x + k = \left( x + a \right)^2\)
Expand the bracket
i.e. \(\scriptsize x^2 – 20x + k = x^2 + 2xa + a^2 \)
Equating coefficient of x
⇒ -20 = 2a
⇒ a = -10
Equating the constant
⇒ k = a2
k = (-10)2
⇒ k = 100
⇒ \(\scriptsize x^2 \: – \: 20x \: + \: k = x^2 \: – \: 20x \: + \: 100 \)
⇒ \( \scriptsize x^2 \: – \: 20x\: + \: 100 = (x \: – \: 10)^2 \)
ii. \(\scriptsize 3x^2 \: + \: 10x \)
Factorise
\( \scriptsize 3 \left ( x^2 \: + \: \normalsize \frac{10x}{3} \right ) \)Let \( \scriptsize x^2 + \normalsize \frac{10x}{3} \scriptsize + k = x^2 + 2xa + a^2 \)
Equating coefficient of x
⇒ \( \frac{10}{3} \scriptsize x = 2xa \)
a = \( \frac{5}{3} \)
Equating constants
⇒ k = \( \left ( \frac{5}{3} \right )^2 = \frac{25}{9} \)
⇒ \( \scriptsize x^2\: +\: \normalsize \frac{10x}{3} \:+ \:\scriptsize k = \scriptsize x^2 \:+ \:\normalsize \frac{10x}{3} \:+\:\frac{25}{9}\)
⇒ \( \scriptsize 3 \left ( \scriptsize x^2 \: +\: \normalsize \frac{10x}{3} \: + \: \frac{25}{9} \right ) \\ = \scriptsize 3 \left (x\: +\: \normalsize \frac{5}{3} \right)^2\)
OR
⇒ \( \scriptsize 3 x^2 \:+ \:10x \: +\: \normalsize \frac{25}{3} \\= \scriptsize 3 \left (x \: + \: \normalsize \frac{5}{3} \right)^2\)
iii. \( \normalsize \frac{1}{4}\scriptsize x^2 \: – \: \normalsize \frac{2}{3}\scriptsize x \)
⇒ \( \frac{1}{4}\scriptsize x^2 \: – \: \normalsize \frac{2}{3} \scriptsize x = \normalsize \frac{1}{4}\left(\scriptsize x^2 \: – \: \normalsize \frac{8}{3} \scriptsize x \right )\)
⇒ Let, \( \: \: \scriptsize x^2 \: – \: \normalsize \frac{8}{3}\scriptsize x \:+\: k = (x \:+\: a)^2 \)
⇒ \(\scriptsize x^2 \: – \: \normalsize \frac{8}{3}\scriptsize x \: + \: k = x^2 \: + \: 2xa \: +\: a^2 \)
Comparing coefficient of x
⇒ \( \frac{-8}{3} \scriptsize x = 2xa \)
⇒ a = \( \frac{-4}{3} \)
Equating constants
k = \( \left ( \frac{-4}{3} \right)^2 = \frac{16}{9} \)
⇒ \(\scriptsize x^2 \: -\: \normalsize \frac{8}{3} \scriptsize x \:+\: k = \scriptsize x^2\: – \:\normalsize \frac{8}{3}\scriptsize x \:+\: \normalsize \frac{16}{9} \)
⇒ \( \frac {1}{4} \left (\scriptsize x^2\: – \: \normalsize \frac{8}{3} \scriptsize x \:+\: \normalsize \frac{16}{9} \right) \scriptsize = \normalsize \frac {1}{4}\left ( \scriptsize x \: – \: \normalsize \frac{4}{3} \right)^2 \)
or
⇒ \(\frac {x^2}{4}\: – \: \frac{2}{3}x \:+\: \frac{4}{9} = \frac {1}{4}\left ( \scriptsize x \:- \: \normalsize \frac{4}{3} \right)^2 \)
iv. \(\scriptsize 4x^2 \: – \: \normalsize \frac{2}{3} \scriptsize x \)
⇒ \( \scriptsize 4 \left (\scriptsize x^2\: -\: \normalsize \frac{1}{6}\scriptsize x \right) = \scriptsize 4 \left (\scriptsize x^2\: -\: \normalsize \frac{x}{6} \right) \)
Let \(\: \scriptsize x^2\: – \: \normalsize \frac{x}{6} \scriptsize+ k = (x \:+ \:a)^2 \)
⇒ \(\: \scriptsize x^2 \:-\: \normalsize \frac{x}{6} \scriptsize\: +\: k = x^2 \:+ \:2xa \:+\: a^2 \)
Equating coefficient of x
⇒ \( \frac{-x}{6} \scriptsize = 2xa \)
a = \( \frac{-1}{12}\)
Equating constants
K = \(\left( \frac{-1}{12}\right)^2 \scriptsize = \normalsize \frac{1}{144}\)
⇒ \(\scriptsize x^2 \:- \:\normalsize \frac{x}{6} \scriptsize \: +\: k = x^2\: – \: \normalsize \frac{x}{6} \scriptsize \:+\: \normalsize \frac{1}{144}\)
⇒ \( \; \scriptsize 4 \left (\scriptsize x^2\: – \: \normalsize \frac{x}{6} \scriptsize \:+\: \normalsize \frac{1}{144} \right) \scriptsize = 4 \left (\scriptsize x \: – \: \normalsize \frac{1}{12} \right )^2\)
or
⇒ \( \scriptsize 4x^2\: – \: \normalsize \frac{2x}{3}\scriptsize \:+\: \normalsize \frac{1}{36}\scriptsize = 4 \left (\scriptsize x \: – \: \normalsize \frac{1}{12} \right )^2\)
Example 5.1.2:
Solve each of the following by completing the square.
i. \( \scriptsize x^2 \:+ \:9x \:+ \:17 = 0\)
ii. \( \scriptsize 3x^2 = \: – \:9x \:+\: 6 \)
iii. \( \scriptsize 5x^2 \:+ \:4x \: – \: 6 = 0 \)
iv. \( \normalsize \frac{3}{4} \scriptsize x^2 \; – \; \normalsize \frac{2}{3} \scriptsize x \; – \; 20 = 0 \)
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