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+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
3 answers
both correct
Thankyou.
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
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
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
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
3 answers