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December 18, 2014

December 18, 2014

Posted by **John** on Tuesday, November 13, 2007 at 6:30pm.

Solve using the square root property:

(x+6)^2 = 4

x+6 = sqrt4 x+6 = -sqrt4

x+6 = 2 x+6 = -2

x+6-6 = 2-6 x+6-6 = -2-6

x = 4 x = -8

Solve by completing the square:

x^2+6x+1=0

x^2+6x+1-1 = 0-1

x^2+6x = -1

x^2+6x+9 = -1+9

(x+3)^2 = -1+9

(x+3)^2 = 8

sqrt(x+3)^2 = sqrt8

x+3 = +-sqrt8

x+3-3 = -3 +-sqrt8

x = -3 +-sqrt8

x = -3 -sqrt8 , x = -3 +sqrt8

x = -3 -2sqrt2 , x = -3 +2sqrt2

- Algebra -
**Reiny**, Tuesday, November 13, 2007 at 6:48pmboth correct

- Algebra -
**John**, Tuesday, November 13, 2007 at 6:53pmThankyou.

Any idea what I am doing wrong in step 7 of this problem?

Solve using the quadratic formula :

x^2-3x=7x-2

x^2-3x-7x+2=7x-7x-2+2

x^2-10x+2=0

x = 10 +- sqrt10^2 - 4(1)(2) / 2(1)

x = 10 +- sqrt100 - 8 / 2

x = 10 +- sqrt92 / 2

x = 10 +- 2sqrt23 /2 (I divided 10by2 and 2by2 and came up with the next step.)

x = 5 +- sqrt23

x = 5 + sqrt23 and x = 5 - sqrt23

- Algebra -
- Algebra -
**Reiny**, Tuesday, November 13, 2007 at 7:30pmother than using brackets in the proper places, your solution is correct

I would write your final lines this way...

x = [10 +- sqrt10^2 - 4(1)(2)] / 2(1)

x = (10 ± √92)/2

= (10 ± 2√23)/2

= 5 ± √23

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