Answer:
1. (x - 3)² = 8
2. (x + 2)² = 3
3. (x + 6)² = [tex]$ \frac{101}{2} $[/tex]
4. (x + 3)² = 27
5. (x + 4)² = 13
6. [tex]$ \bigg( x - \frac{15}{9} \bigg) ^2 = \frac{261}{81} = \frac{29}{9} $[/tex]
Step-by-step explanation:
Completion of Square: [tex]$ (x - a) ^2 = x^2 - 2ax + a^2 $[/tex]
In the following problems the terms in the RHS of the above equation may be missing. We balance the equation. Simplify it and re write it in terms of LHS.
1. x² - 6x + 1 = 0
Taking the constant term to the other side, we get:
x² - 6x = - 1
⇒ x² - 2(3)x = -1
⇒ x² -2(3)x + 9 = - 1 + 9 [Adding 9 to both the sides]
⇒ x² -2(3)x + 3² = 8
⇒ (x - 3)² = 8 is the answer.
2. 3x² + 12x + 3 = 0
Note that the co-effecient of x² is not 1. We make it 1, by dividing the whole equation by 3. And then proceed like the previous problem.
3x² + 12x = -3
Dividing by 3 through out, x² + 4x = - 1
⇒ x² + 2(2) + 4 = -1 + 4
⇒ x² +2(2) + 2² = 3
⇒ (x + 2)² = 3 is the answer.
3. 2x² + 24x = 29
x² + 12x = [tex]$ \frac{29}{2} $[/tex]
⇒ x² + 2(6)x + 36 = [tex]$ \frac{29}{2} $[/tex] + 36
⇒ x² + 2(6)x + 6² = [tex]$ \frac{29 + 72}{2} $[/tex]
⇒ (x + 6)² = [tex]$ \frac{101}{2} $[/tex] is the answer.
4. x² + 6x - 18 = 0
x² + 6x = 18
⇒ x² + 2(3)x = 18
⇒ x² + 2(3)x + 9 = 18 + 9
⇒ x² + 2(3)x + 3² = 27
⇒ (x + 3)² = 27 is the answer.
5. x² + 8x + 3 = 0
x² + 8x = -3
⇒ x² + 2(4)x = -3
⇒ x² + 2(4)x + 16 = - 3 + 16
⇒ x² + 2(4)x + 16 = 13
⇒ (x + 4)² = 13 is the answer.
6. 9x² - 30x + 6 = 0
9x² - 30x = - 6
⇒ x² [tex]$ - \frac{30}{9} $[/tex] x = - 6
[tex]$ \implies x^2 -2 \bigg( \frac{15}{9} \bigg )x + \frac{225}{81} = - 6 + \frac{225}{81} $[/tex]
[tex]$ \implies x^2 - 2\bigg( \frac{15}{9} \bigg ) x + \bigg ( \frac{15}{9} \bigg ) ^2 = \frac{261}{81} $[/tex]
[tex]$ \bigg( x - \frac{15}{9} \bigg) ^2 = \frac{261}{81} = \frac{29}{9} $[/tex] is the answer.
