Question

Solve the quadratic equation by completing the square.
x2 + 6x + 7 = 0

Answer 1

x1 = -1.59

x2 = -4.41

Explanation:

The quadratic equation is of the form

`ax^(2)+bx+c=0`

In the given case, b = 6

We apply the formula

`((b)/(2))^(2)=((6)/(2))^(2)=3^(2)=9`

`x^(2)+6x+7+9-9=0`

`x^(2)+6x+9+7-9=0`

`(x^(2)+6x+9)-2=0`

`(x+3)(x-3)=2`

`(x+3)^(2)=2`

Applying square root to both parts of the equation, we get,

`\sqrt{(x+3)^(2)} =√(2)`

`x+3=√(2)`

`x=√(2)-3`

x = ± 1.41 - 3

"x" then has two values:

x1 = 1.41 - 3 = -1.59

x2 = -1.41 - 3 = -4.41

Hope this helps!

Answer 2

from the equation x^2 - 6x + 7 = 0 , b is equal to -6. The form should become y = (x-(b/2)^2) + c. (b/2)^2 is equal to 9. Hence,

0 = (x - 3)^2 -9 + 7
0 = (x - 3)^2 -2
2 = (x - 3)^2

x should be equal to -1.5858 and -4.4142