Best solutions to - "How do I solve for x algebraically?x2(x2−1)x+3=12"

Kale Bright

Answered question

2022-04-17

How do I solve for x algebraically?

\frac{x^{2} (x^{2} - 1)}{x + 3} = 12

ruseducatives1t03

Beginner2022-04-18Added 8 answers

Rearranging the equation, we get

x^{2} (x^{2} - 1) = 12 (x + 3)

x^{4} - x^{2} - 12 x - 36 = 0

First we search for integer roots. The integer root must divide 36. Hence, the possible integer options are

\pm 1, \pm 2, \pm 3

. Checking these 6 options, give us

x = - 2

and

x = 3

. Hence,

x^{4} - x^{2} - 12 x - 36 = (x + 2) (x - 3) (x^{2} + a x + b)

Comparing coefficients, we get

a = 1

and

b = 6

. Solving the quadratic, gives the other roots as

x^{2} + x + 6 = 0 \Rightarrow {(x + \frac{1}{2})}^{2} + 6 - \frac{1}{4} = 0 \Rightarrow x = - \frac{1}{2} \pm i \frac{\sqrt{23}}{2}

cinereod3am

Beginner2022-04-19Added 10 answers

Multiplying both sides by

x + 3

gives

x^{2} (x^{2} - 1) = 12 (x + 3)

. This can be re-written as

x^{4} - x^{2} - 12 x - 36 = 0

. A few guesses shows that 3 and -2 are solutions, so you can factor these out to get a quadratic.
Explicitly, the equation becomes

(x - 3) (x + 2) (x^{2} + x + 6) = 0

, hence the solutions are

- 3, - 2

, and

\frac{1}{2} (- 1 \pm i \sqrt{23}

(And, of course, we check that none of these are

- 3

)

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