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# I hate intermediate algebra....

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What number is 8 more than its recipicol?

Jun 1, 2019

#1
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Solve for x:
8 + 1/x = x

Bring 8 + 1/x together using the common denominator x:
(8 x + 1)/x = x

Multiply both sides by x:
8 x + 1 = x^2

Subtract x^2 from both sides:
-x^2 + 8 x + 1 = 0

Multiply both sides by -1:
x^2 - 8 x - 1 = 0

x^2 - 8 x = 1

x^2 - 8 x + 16 = 17

Write the left hand side as a square:
(x - 4)^2 = 17

Take the square root of both sides:
x - 4 = sqrt(17) or x - 4 = -sqrt(17)

x = 4 + sqrt(17) or x - 4 = -sqrt(17)

x = 4 + sqrt(17)         or                x = 4 - sqrt(17)

Jun 1, 2019
#2
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x = 1/x + 8     is what is being asked   multiply by x

x^2 = 1+8x     re-arrange

x^2-8x-1=0     use quadratic formula   a = 1 b = -8  c = -1

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

{8+-sqrt(64+4(1)(1))   } / 2

4 +- sqrt(17*4)/2

4+- 2(sqrt17)/2

4 +- sqrt 17

Jun 2, 2019