Algebra
posted by John
Two more I would like to check my work on:
Solve using the square root property:
(x+6)^2 = 4
x+6 = sqrt4 x+6 = sqrt4
x+6 = 2 x+6 = 2
x+66 = 26 x+66 = 26
x = 4 x = 8
Solve by completing the square:
x^2+6x+1=0
x^2+6x+11 = 01
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+33 = 3 +sqrt8
x = 3 +sqrt8
x = 3 sqrt8 , x = 3 +sqrt8
x = 3 2sqrt2 , x = 3 +2sqrt2

Reiny
both correct

John
Thankyou.
Any idea what I am doing wrong in step 7 of this problem?
Solve using the quadratic formula :
x^23x=7x2
x^23x7x+2=7x7x2+2
x^210x+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 
Reiny
other 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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