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# Quadratic equation formula, where did I go wrong?

The question is, find the value of u using the quadratic formula

u + 1 + 3/(u + 1) = 8

u(u + 1) + u + 1 + 3 = 8(u + 1)

u^2 + u + u + 1 + 3 = 8u + 8

u^2 + 2u + 4 = 8u + 8

....subtracting 2u and 4 from both sides...

u^2 = 6u + 4

..subtracting 6u and 4 from both sides...

u^2 - 6u - 4 = 0

Using the formula....

[-(-6) +/- √(-6)^2 - 4 x 1 x (-4)] / 2 x 1

[6 +/- √36 +16] / 2

[6 +/- √52] / 2

so...

(6 + 7.211) / 2 = 6.61 = u

or (6 - 7.211) / 2 = -0.61 = u

But this is wrong! Where did I go wrong? Thanks

### 5 Answers

- billrussell42Lv 72 years agoFavorite Answer
I changed the negative answer to 4 places, and then it passes OK.

u = 6.61, – 0.6055

just check the answers

check

u + 1 + 3/(u + 1) = 8

7.61 + 3/7.61 = 8

8.00 = 8

ok

– 0.6055 + 1 + 3/(– 0.6055 + 1) = 8

0.3945 + 3/0.3945 = 8

7.999=8

ok

- King LeoLv 72 years ago
.

u + 1 + 3/(u + 1) = 8

multiply both sides of the equation by u + 1;

(u + 1)(u + 1) + 3 = 8(u + 1)

then simplify

u² + 2u + 1 + 3 = 8u + 8

u² - 6u - 4 = 0

complete the square or use the quadratic formula

(u - 3)² = 4 + 9

(u - 3)² = 13

u - 3 = ±√13

u = 3 ± √13

━━━━━

- Anonymous2 years ago
You didn't do anything wrong. Plugging back into the original equation, you'll see that it checks out. Perhaps they want the exact radical solution, or perhaps they want you to round to a certain decimal place.

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- 2 years ago
(u + 1) + 3 / (u + 1) = 8

I'd let u + 1 = k and go from there

k + 3/k = 8

k^2 + 3 = 8k

k^2 - 8k + 3 = 0

k = (8 +/- sqrt(64 - 12)) / 2

k = (8 +/- sqrt(52)) / 2

k = (8 +/- 2 * sqrt(13)) / 2

k = 4 +/- sqrt(13)

u + 1 = 4 +/- sqrt(13)

u = 3 +/- sqrt(13)

u = 3 + sqrt(13) , 3 - sqrt(13)

You went wrong in your final step. Maybe you shouldn't round it off. Just give it to them in a radical form.