I can't figure out how to work out this problem... 12/√(5x+6)=√(2x+5)?
2009-06-03 8:40 am
My brain is totally fried after trying to figure this out. I can't comprehend what my book is explaining to me, as far as the steps to take goes. Does anyone have any helpful tips on how I should start this off? I'd greatly appreciate it.
回答 (2)
2009-06-03 8:49 am
✔ 最佳答案
Square both sides to eliminate the radicals:(12/√(5x+6))^2=(√(2x+5))^2
144/(5x+6) = (2x+5).
Clear the fractions:
144 = (5x+6)(2x+ 5)
144 = 10x^2 + 37x + 30
10x^2 + 37x - 114 = 0.
(10x + 57)(x - 2) = 0.
So, x = -57/10 or x = 2.
Double check to see these are solutions; do this any time you square both sides of an equation!
We see that x = -57/10 can't work, since we can't take the square root of a negative number. This is an "extraneous solution".
On the other hand, 12/√(5*2+6)=√(2*2+5)
12/4 = 3.
So, x = 2 is the only solution.
2009-06-03 11:45 am
12/â(5x + 6) = â(2x + 5)
12 = â[(5x + 6)(2x + 5)]
12^2 = 5x*2x + 6*2x + 5x*5 + 6*5
144 = 10x^2 + 12x + 25x + 30
10x^2 + 37x + 30 - 144 = 0
10x^2 + 37x - 114 = 0
10x^2 + 57x - 20x - 114 = 0
(10x^2 + 57x) - (20x + 114) = 0
x(10x + 57) - 2(10x + 57) = 0
(10x + 57)(x - 2) = 0
10x + 57 = 0
10x = -57
x = -57/10 (-5.7)
x - 2 = 0
x = 2
â´ x = -57/10, (-5.7), 2
12 = â[(5x + 6)(2x + 5)]
12^2 = 5x*2x + 6*2x + 5x*5 + 6*5
144 = 10x^2 + 12x + 25x + 30
10x^2 + 37x + 30 - 144 = 0
10x^2 + 37x - 114 = 0
10x^2 + 57x - 20x - 114 = 0
(10x^2 + 57x) - (20x + 114) = 0
x(10x + 57) - 2(10x + 57) = 0
(10x + 57)(x - 2) = 0
10x + 57 = 0
10x = -57
x = -57/10 (-5.7)
x - 2 = 0
x = 2
â´ x = -57/10, (-5.7), 2
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