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I am to solve for x using the quadratic formula:

$$4+\frac{1}{x}-\frac{1}{x^2}=0$$

The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$

Here's my working:

Start with $$4+\frac{1}{x}-\frac{1}{x^2}=0$$

Rearranging into standard form:

$$-\frac{1}{x^2}+\frac{1}{x}+4=0$$

Multiply by $-1$ to get a positive leading coefficient $a$:

$$\frac{1}{x^2}-\frac{1}{x}-4=0$$

I'm not sure how to determine my inputs $a,b$ and $c$ with these fractions but I guess $a=\dfrac{1}{x^2}$, $b=\dfrac{1}{x}$ and $c=-4$.

Plugging into quadratic function:

$$x = \frac{-\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$

I find this challenging due to the coefficients $a$ and $b$ being fractions.

How can I apply the quadratic formula to $4+\dfrac{1}{x}-\dfrac{1}{x^2}=0$ to arrive at $\dfrac{-1\pm\sqrt{17}}{8}$?

JacobCheverie
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Doug Fir
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    Try multiplying the equation by $x^2,$ and then you should have a quadratic equation in a familiar form – J. W. Tanner May 03 '19 at 11:57
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    Indeed, you need to multiply with $x^2$ to get somewhere. – Wuestenfux May 03 '19 at 11:57
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    You actually solved the quadratic equation $$\dfrac{t^2}{x^2}+\dfrac tx-4$$ for $t$. –  May 03 '19 at 12:01
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    As an aside, I'd like to say that this is a well written question in my opinion. It clearly shows an attempt and that you have some understanding of the problem and where your attempt got stuck. Good luck in your studies and I hope you continue to post, getting help here and eventually returning the favor by answering other user's questions. – JMoravitz May 03 '19 at 12:03
  • If you choose the coefficients of your quadratic equation to be $1/x^2$ and so on, then what is the unknown variable in your equation? – amd May 03 '19 at 18:58

6 Answers6

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Multiplying both sides by $x^2$ will result in

$$4x^2+x-1=0$$

Now, with $a=4, b=1, c=-1$ use the quadratic formula and let us know what you get.

JacobCheverie
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    Thanks, this tip allowed me to arrive at my textbook's solution. – Doug Fir May 03 '19 at 12:13
  • Just to make this super clear.. This is correct as long as you have to calculate the roots (zeros) of the equation. For any other operations you will have to use the pervious form – 44yu5h Apr 25 '23 at 06:29
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Note that$$4+\frac1x-\frac1{x^2}=\frac1{x^2}\left(4x^2+x-1\right).$$So, solve the equation $4x^2+x-1=0$.

José Carlos Santos
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Other answers followed my suggestion in the comments.

Here's an alternative:

Let $z=\dfrac1x.$ Then we have $-z^2+z+4=0$, so, using the quadratic formula, $z=\dfrac{-1\pm\sqrt{17}}{-2}.$

Therefore $x=\dfrac1z=\dfrac{-2}{-1\pm\sqrt{17}}=\dfrac{2(-1\mp\sqrt{17})}{16}.$

J. W. Tanner
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From

$$4+\left(\frac1x\right)-\left(\frac1x\right)^2=0$$

using the standard formula blindly,

$$\frac1x=\frac{1\pm\sqrt{17}}2$$

and obviously

$$x=\frac2{1\pm\sqrt{17}}.$$

Though this seems to contradict the expected answer, consider

$$\frac2{1\pm\sqrt{17}}=\frac{2(1\mp\sqrt{17})}{1-(\sqrt{17})^2}=\frac{-1\pm\sqrt{17}}8.$$

  • As an edit, you might include the fact that setting $u=\frac1x$ makes the quadratic equation easier to see (thus the formula easier to 'blindly' apply). – Clayton May 03 '19 at 12:07
  • @Clayton: I took the care to parenthesize the square rather than leaving the squared denominator. This is enough for me. I will parenthesize the other fraction. Please note that $\frac1x$ corresponds to a $u$ in the Badfaithian alphabet. :) –  May 03 '19 at 12:08
  • You and I had the same idea simultaneously – J. W. Tanner May 03 '19 at 12:14
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    @J.W.Tanner: yep, you can always solve a polynomial two ways, by considering the inverse of the unknown. That causes a swap of the coefficients. –  May 03 '19 at 12:19
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Note that $$4+\frac1x-\frac1{x^2}=4+\frac1x-\left(\frac1x\right)^2,$$ so making the substitution $y=\frac1x,$ we obtain the quadratic $$-y^2+y+4=0,$$ which should be more familiar. Solve for $y,$ and since neither solution for $y$ should be equal to $0,$ use $x=\frac1y$ to solve for $x.$

Cameron Buie
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None of the other answers so far has really addressed the error that you made in your attempt. After rearranging the original equation into $$\frac{1}{x^2}-\frac{1}{x}-4=0,$$ you then decided that $a=1/x^2$, $b=-1/x$ and $c=-4$. Substituting these names for the corresponding values in this equation gives you $$a+b+c=0.$$ This is no longer a quadratic equation—it doesn’t even have an unknown to solve for!

Instead, as other answers explain, you need to either multiply by $x^2$ to eliminate $x$ from all of the denominators, or introduce a new variable such as $y=1/x$. Either approach will give you an equation that looks more like one you’re used to. Remember to reject $x=0$ if it comes up as a solution to the modified equation (it won’t).

amd
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