Solve the rational equation below
1. z^2/8 -7z/8 =1
2. 2n^2/2 - 5n/6
2007-12-28 05:37:42 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
sorry I wrote this one wrong
2. 2n^2/2 - 5n/6 -2 =0
2007-12-28 05:38:34 · update #1
z^2/8 -7z/8 =1
Find the lowest comment denominator (LCD) for all three terms, which would be 8, so we get
z^2 -7z = 8
z^2 -7z -8 =0
(z-8)(z+1)=0
z=8,-1
2n^2/2 - 5n/6 -2 =0
Find the LCD again for the terms, which would be 6, so we get
6n^2 - 5n -12 =0
Using quadratic, we get n= 1.89, -1.058
2007-12-28 05:45:17 · answer #1 · answered by ¿ /\/ 馬 ? 7 · 0⤊ 0⤋
1. Since both of the problems before the equal sign have the same denominator, you can multiply both sides by 8 to remove the fraction. So, z^2-7z=8
Move the 8 on over to the left:
z^2 - 7z - 8 = 0
Go ahead and factor that:
(z-8)(z+1) = z^2-7z-8
Z = 8 AND Z = -1
2. It looks like it'd work out to be 6n^2-5n-12 but that doesn't have any rational zeros so I'm unsure.
2007-12-28 05:48:54 · answer #2 · answered by heystevenn 4 · 0⤊ 0⤋
1. z^2/8 -7z/8 =1
z^2 -7z -8 = 0
(z-8)(z+1) = 0
z = 8 or -1
2. 2n^2/2 - 5n/6 -2 = 0
n = [5 +/- sqrt(41)]/4
2007-12-28 05:48:22 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1. z^2/8 -7z/8 =1
z^2 -7z -8 = 0
(z-8)(z+1) = 0
z = 8 or -1
2. 2n^2/2 - 5n/6 -2 = 0
6n^2 -5n -12= 0
n = [5 +/- sqrt(5^2 -4(6)(-12))]/(2*6)
n = [5 +/- sqrt(313)]12
2007-12-28 05:45:04 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋