Math  /  Algebra

Question150 Solve: 14r=r2+5r+6r12r\frac{1}{4 r}=\frac{r^{2}+5 r+6}{r}-\frac{1}{2 r}

Studdy Solution

STEP 1

1. The equation involves rational expressions.
2. We will need to find a common denominator to combine the terms on the right side.
3. We will solve for r r by simplifying and solving the resulting equation.

STEP 2

1. Simplify the right side of the equation.
2. Combine the terms on the right side.
3. Solve for r r .

STEP 3

First, simplify the right side of the equation by breaking it into two separate fractions:
r2+5r+6r12r \frac{r^2 + 5r + 6}{r} - \frac{1}{2r}
The first fraction can be simplified by dividing each term in the numerator by r r :
r2r+5rr+6r=r+5+6r \frac{r^2}{r} + \frac{5r}{r} + \frac{6}{r} = r + 5 + \frac{6}{r}

STEP 4

Combine the terms on the right side by finding a common denominator. The common denominator for the terms is 2r 2r :
r+5+6r12r r + 5 + \frac{6}{r} - \frac{1}{2r}
Convert r+5 r + 5 into a fraction with denominator 2r 2r :
2r(r+5)2r=2r2+10r2r \frac{2r(r + 5)}{2r} = \frac{2r^2 + 10r}{2r}
Now, combine all terms over the common denominator 2r 2r :
2r2+10r+1212r=2r2+10r+112r \frac{2r^2 + 10r + 12 - 1}{2r} = \frac{2r^2 + 10r + 11}{2r}

STEP 5

Set the left side equal to the simplified right side and solve for r r :
14r=2r2+10r+112r \frac{1}{4r} = \frac{2r^2 + 10r + 11}{2r}
Cross-multiply to eliminate the fractions:
12r=4r(2r2+10r+11) 1 \cdot 2r = 4r(2r^2 + 10r + 11)
Simplify and solve the resulting equation:
2r=8r3+40r2+44r 2r = 8r^3 + 40r^2 + 44r
Rearrange the equation to set it to zero:
8r3+40r2+44r2r=0 8r^3 + 40r^2 + 44r - 2r = 0 8r3+40r2+42r=0 8r^3 + 40r^2 + 42r = 0
Factor out the greatest common factor, which is 2r 2r :
2r(4r2+20r+21)=0 2r(4r^2 + 20r + 21) = 0
Set each factor equal to zero:
1. 2r=0 2r = 0 implies r=0 r = 0
2. Solve 4r2+20r+21=0 4r^2 + 20r + 21 = 0 using the quadratic formula:

r=b±b24ac2a r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=4 a = 4 , b=20 b = 20 , and c=21 c = 21 .
r=20±202442124 r = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 4 \cdot 21}}{2 \cdot 4} r=20±4003368 r = \frac{-20 \pm \sqrt{400 - 336}}{8} r=20±648 r = \frac{-20 \pm \sqrt{64}}{8} r=20±88 r = \frac{-20 \pm 8}{8}
Calculate the roots:
r=20+88=128=32 r = \frac{-20 + 8}{8} = \frac{-12}{8} = -\frac{3}{2} r=2088=288=72 r = \frac{-20 - 8}{8} = \frac{-28}{8} = -\frac{7}{2}
The possible solutions for r r are r=0 r = 0 , r=32 r = -\frac{3}{2} , and r=72 r = -\frac{7}{2} .
The solutions for r r are:
r=0,r=32,r=72 r = 0, \, r = -\frac{3}{2}, \, r = -\frac{7}{2}

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