# what are the roots of this rational equation? (3/x) + (4/(x+3)) = 1?

### 3 Answers

- billrussell42Lv 79 months agoFavourite answer
simplify it first

(3/x) + (4/(x+3)) = 1

multiply by x(x+3)

3(x+3) + 4x = x(x+3)

3x + 9 + 4x – x² – 3x = 0

– x² + 4x + 9 = 0

x² – 4x – 9 = 0

quadratic equation:

to solve ax² + bx + c = 0

x = [–b ± √(b²–4ac)] / 2a

x = [4 ± √(16+36)] / 2

x = [4 ± √52] / 2

x = 2 ± √13

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- KrishnamurthyLv 79 months ago
(3/x) + (4/(x + 3)) = 1

3(x + 3) + 4x = x(x + 3)

x^2 - 4x - 9 = 0

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- 9 months ago
If we get answers of x = 0 or x = -3, then they are extraneous, because we'd have 0s in a denominator

3/x + 4/(x + 3) = 1

(3 * (x + 3) + 4 * x) / (x * (x + 3)) = 1

3x + 9 + 4x = x * (x + 3)

7x + 9 = x^2 + 3x

0 = x^2 - 4x - 9

x = (4 +/- sqrt(16 + 36)) / 2

x = (4 +/- sqrt(52)) / 2

x = (4 +/- 2 * sqrt(13)) / 2

x = 2 +/- sqrt(13)

Test

3/(2 + sqrt(13)) + 4/(2 + sqrt(13) + 3) =>

3 * (sqrt(13) - 2) / (13 - 4) + 4 / (sqrt(13) + 5) =>

3 * (sqrt(13) - 2) / 9 + 4 * (sqrt(13) - 5) / (13 - 25) =>

(sqrt(13) - 2) / 3 + 4 * (sqrt(13) - 5) / (-12) =>

(sqrt(13) - 2) / 3 - (sqrt(13) - 5) / 3 =>

(sqrt(13) - 2 - sqrt(13) + 5) / 3 =>

3/3 =>

1

3 / (2 - sqrt(13)) + 4 / (2 - sqrt(13) + 3) =>

3 * (2 + sqrt(13)) / (4 - 13) + 4 * (5 + sqrt(13)) / (25 - 13) =>

3 * (2 + sqrt(13)) / (-9) + 4 * (5 + sqrt(13)) / 12 =>

(5 + sqrt(13)) / 3 - (2 + sqrt(13)) / 3 =>

(5 + sqrt(13) - 2 - sqrt(13)) / 3 =>

3 / 3 =>

1

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