Question

Solve each quadratic equation by completing the square. 6. x2 + 2x = 8 7. x2 - 6x = 16 8. x2 - 18x = 19 9. x2 + 3x = 3 10. x2 + 7x = 8

Answer 1

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 $+ (\frac{2}{2} )^2 = 8 + ( \frac{2}{2})^2$
x² + 2x + 1 = 8 + 1
( x + 1 )( x + 1 ) = 9
( x + 1 )² = 9
$\sqrt{( x + 1 )^2} = \sqrt{9}$
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 + $(\frac{6}{2})^2 = 16 + (\frac{6}{2})^2$ 
x² - 6x + 9 = 16 + 9
(x - 3)² = 25
$\sqrt{(x-3)^2} = \sqrt{25}$
x - 3 = + 5 or x - 3 = - 5
   x = 8      or       x = - 2

8. x² - 18x = 19

x² - 18x + $( \frac{18}{2} )^2 = 19 + ( \frac{18}{2})^2$
x² - 18x + 81 = 19 + 81
( x - 9 )( x - 9 ) = 100
( x - 9 )² = 100
$\sqrt{(x-9)^2} = \sqrt{100}$
x - 9 = + 10 or x - 9 = -10
   x = 19      or      x = - 1

9. x² + 3x = 3

x² + 3x + $( \frac{3}{2} )^2 = 3 + (\frac{3}{2} )$
$x^2 + 3x + \frac{9}{4} = 3 + \frac{9}{4}$
$x^{2} + 3x + \frac{9}{4} = \frac{21}{4}$
$(x^2 + \frac{3}{2} ) ( x^2 + \frac{3}{2} ) = \frac{21}{4}$
$( x^2 + \frac{3}{2} )^2 = \frac{21}{4}$
$\sqrt{ x^2 + \frac{3}{2} } = \sqrt{ \frac{21}{4} }$ 
x +  $\frac{3}{2}$ = + $\frac{ \sqrt{21} }{2}$ or x +$\frac{3}{2} = - \frac{ \sqrt{21} }{2}$
x = $\frac{-3+ \sqrt{21} }{2}$ or x = $\frac{-3 - \sqrt{21}}{2}$