Solve the equation: $$\dfrac{7}{x+\sqrt{x+5}}+\dfrac{7}{x-\sqrt{x+5}}=8.$$ I am not sure how to approach the problem. Should we first determine the domain? I think we can also check for every value we get for $x$ if the expression is defined, or not. The answers are $x=-\dfrac{5}{4}$ and $x=4.$ The numerators are equal and we also have resemblance in the denomitators. I am not sure how to use that. Thank you in advance! Happy holidays!
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asked Dec 21, 2020 at 16:53
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1$\begingroup$ Think about getting both fractions in one bigger fraction, then it will simplify a lot. $\endgroup$– jojoboCommented Dec 21, 2020 at 16:57
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1$\begingroup$ I suggest rationalizing the denominators. $\endgroup$– VishuCommented Dec 21, 2020 at 17:00
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3 Answers
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$$\dfrac{7}{x+\sqrt{x+5}}+\dfrac{7}{x-\sqrt{x+5}} = \frac{7(x-\sqrt{x+5})}{x^2-x-5} +\frac{7(x+\sqrt{x+5})}{x^2-x-5} \\ = \frac{14x}{x^2-x-5}$$ So the equation is equivalent to $$\frac{14x}{x^2-x-5} =8 \\ 14x =8x^2-8x-40 \\ 8x^2-22x-40=0$$
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$\begingroup$ Thank you for the response! Should I substitute in $\dfrac{14x}{x^2-x-5}=8$ to see if the expression is defined? Is it enough just to substitute in the original equation? I mean are we sure if the original equation is defined, then the mentioned by me is also defined? $\endgroup$ Commented Dec 21, 2020 at 19:13
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1$\begingroup$ Because at each step we are saying “if this is true, then <new equation> has to be true.” $\endgroup$– VishuCommented Dec 21, 2020 at 19:18
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1$\begingroup$ Okay! I am still not sure I understand it, but I am trying. :) $\endgroup$ Commented Dec 21, 2020 at 19:24
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1$\begingroup$ Think of when an expression won’t be defined. Now if there’s some discrepancy in a middle step, wouldn’t the step before it also have the same discrepancy? $\endgroup$– VishuCommented Dec 21, 2020 at 19:27
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1$\begingroup$ In this context, I mean something that would make an expression undefined. $\endgroup$– VishuCommented Dec 21, 2020 at 19:29
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\begin{align*} \frac{7}{x + \sqrt{x+5}} + \frac{7}{x - \sqrt{x + 5}} = 8 & \Rightarrow \frac{14x}{x^{2} - x - 5} = 8\\\\ & \Rightarrow 8x^{2} - 22x - 40 = 0\\\\ & \Rightarrow 4x^{2} - 11x - 20 = 0\\\\ & \Rightarrow \left(x = - \frac{5}{4}\right)\vee(x = 4) \end{align*}
Substituting in the original equation, we conclude that they are solutions indeed.
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Note that:
$$\frac{7}{x+\sqrt{x+5}}+\frac{7}{x-\sqrt{x+5}}=\frac{14x}{x^2-x-5}$$
answered Dec 21, 2020 at 17:00