Lets get started :)
These questions have asked us to solve by completing the square.
How do we? I have attached a picture, which will explain
6. x² + 2x = 8
→ b is the coefficient of x, which is 2
→ We take half of 2 and square it. Then, we add it to either side
x² + 2x [tex] + (\frac{2}{2} )^2 = 8 + ( \frac{2}{2})^2 [/tex]
x² + 2x + 1 = 8 + 1
( x + 1 )( x + 1 ) = 9
( x + 1 )² = 9
[tex] \sqrt{( x + 1 )^2} = \sqrt{9} [/tex]
x + 1 = + 3 or x + 1 = - 3
x = 2 or x = - 4
7. x² - 6x = 16
→ We do the same thing we did in the previous question
x² - 6x + [tex] (\frac{6}{2})^2 = 16 + (\frac{6}{2})^2[/tex]
x² - 6x + 9 = 16 + 9
(x - 3)² = 25
[tex] \sqrt{(x-3)^2} = \sqrt{25} [/tex]
x - 3 = + 5 or x - 3 = - 5
x = 8 or x = - 2
8. x² - 18x = 19
x² - 18x + [tex]( \frac{18}{2} )^2 = 19 + ( \frac{18}{2})^2 [/tex]
x² - 18x + 81 = 19 + 81
( x - 9 )( x - 9 ) = 100
( x - 9 )² = 100
[tex] \sqrt{(x-9)^2} = \sqrt{100} [/tex]
x - 9 = + 10 or x - 9 = -10
x = 19 or x = - 1
9. x² + 3x = 3
x² + 3x + [tex]( \frac{3}{2} )^2 = 3 + (\frac{3}{2} )[/tex]
[tex]x^2 + 3x + \frac{9}{4} = 3 + \frac{9}{4} [/tex]
[tex] x^{2} + 3x + \frac{9}{4} = \frac{21}{4} [/tex]
[tex](x^2 + \frac{3}{2} ) ( x^2 + \frac{3}{2} ) = \frac{21}{4} [/tex]
[tex]( x^2 + \frac{3}{2} )^2 = \frac{21}{4} [/tex]
[tex] \sqrt{ x^2 + \frac{3}{2} } = \sqrt{ \frac{21}{4} } [/tex]
x + [tex] \frac{3}{2} [/tex] = + [tex] \frac{ \sqrt{21} }{2} [/tex] or x +[tex] \frac{3}{2} = - \frac{ \sqrt{21} }{2} [/tex]
x = [tex] \frac{-3+ \sqrt{21} }{2} [/tex] or x = [tex] \frac{-3 - \sqrt{21}}{2} [/tex]
