2007-10-26 01:11:50 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
sorry pro, and falzoon and that other guy, danng, u guys are good
2007-10-26 01:31:55 · update #1
oooohhh
(2+(root)2x+5)^2=4 +4 (root)2x+5+2x-5
I mistakenly put 8 +4 ect...
2007-10-26 02:20:08 · update #2
sqrt(4x - 3) = 2 + sqrt(2x - 5)
Square both sides :
4x - 3 = 4 + 4*sqrt(2x - 5) + 2x - 5
Simplify and rearrange :
4*sqrt(2x - 5) = 2x - 2
Square both sides :
16(2x - 5) = 4x^2 - 8x + 4
Expand :
32x - 80 = 4x^2 - 8x + 4
Simplify and rearrange :
4x^2 - 40x + 84 = 0
Divide through by 4 :
x^2 - 10x + 21 = 0
Factorise :
(x - 3)(x - 7) = 0
Therefore, x = 3 or 7.
Check by substituting each into the original equation :
x = 3 implies sqrt(9) = 2 + sqrt(1), which is correct.
x = 7 implies sqrt(25) = 2 + sqrt(9), which is correct.
2007-10-26 01:28:59 · answer #1 · answered by falzoon 7 · 0⤊ 0⤋
flow the sqrt(2x+50) to the different part by utilising subtracting it 2 sqrt (x+6) = 5 - sqrt(2x+50) sq. the two facets (the right part is FOIL) 4 (x+6) = (5 - sqrt(2x+50))(5 - sqrt(2x+50)) 4x + 24 = 25 - 10 sqrt (2x+50) + 2x + 50 flow the ten sqrt (2x+50) to the left and the 4x + 24 to the right by utilising including the different 10 sqrt (2x+50) = -2x + 51 sq. the two facets lower back a hundred (2x+50) = (-2x + 51)(-2x + 51) try this out, make it equivalent to 0, and resolve by utilising quadratic formula or if accessible FOIL you may desire to do it this manner when you consider which you may not sq. the two facets if issues are further. it is you may not in basic terms sq. each component in the unique difficulty and get 2x+50 + 4(x+6) = 25
2016-10-14 02:27:19 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Squaring both sides we get
4x-3 = 4+(2x-5)+2*2*sqrt(2x-5)
=> 4x-3 = 2x-1+4*sqrt(2x-5)
=> 4x-2x-3+1 = 4*sqrt(2x-5)
=> 2x-2 = 4*sqrt(2x-5)
=> x-1 = 2*sqrt(2x-5)
squaring both sides once again
x^2 -2x+1= 4(2x-5)
=> x^2-10x+21 = 0
=> x^2-7x-3x+21 = 0
=> (x^2-7x)-(3x-21) = 0
=> x(x-7)-3(x-7) = 0
=> (x-7)(x-3) =0
=> x-7 = 0 or x-3 = 0
=> x = 7 or x = 3
2007-10-26 01:30:49 · answer #3 · answered by sulinderkumarsharma 2 · 0⤊ 0⤋
Transfer the sq.root terms on one side of the equation and the rest on the other side. Then square both sides. You will need to repeat the process twice.
4x -3 +2x-5 -2sqrt(8x^2 -26x +15) =4
6x - 8 -4=2sqrt(8x^2 -26x +15)
6x - 12= 2sqrt(8x^2 -26x +15)
3x -6=sqrt(8x^2 -26x +15)
square both sides again
9x^2 +36 - 36x=(8x^2 - 26x +15)
x^2 - 10x +21=0
solve using quadratic equation formula:
x=7 or 3
2007-10-26 01:48:19 · answer #4 · answered by vcs7578 5 · 0⤊ 0⤋
sqrt(4x-3)= 2+sqrt(2x-5)
1) x>= 5/2
square both sides
4x-3 = 4+2x-5 +4 sqrt(2x-5)
2x-2= 4 sqrt(2x-5)
x-1= 2sqrt(2x-5) square again
x^2-2x+1 = 8x-20
x^2-10x+21 =0 x= ((10+-sqrt(100-84))/2
so x= 7 and x= 3
2007-10-26 01:29:24 · answer #5 · answered by santmann2002 7 · 0⤊ 0⤋
answer is 7 and 3
2007-10-26 01:27:17 · answer #6 · answered by programhelp 2 · 1⤊ 0⤋