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solve $$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} $$

$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} \Rightarrow $$ $$ 5x+19 = (x+7) + 4\sqrt{x-5}\sqrt{x+7} + (x+5) \Rightarrow $$ $$ 3x + 17 = 4\sqrt{x-5}\sqrt{x+7} \Rightarrow $$ $$ 9x^2 + 102x + 289 = 16(x+7)(x-5) \Rightarrow $$ $$ 9x^2 + 102x + 289 = 16(x^2+ 2x - 35) \Rightarrow $$ $$ 7x^2 - 70x - 849 = 0 \Rightarrow $$ $$ b^2 - 4ac = (-70)^2 - 4 \cdot 7 \cdot (-849) = 28672=2^{12}\cdot7 $$

then I calculate the solution using the discriminant as $$ 5 + 32\frac{\sqrt7}7 $$ and $$ 5 - 32\frac{\sqrt7}7 $$

but when I plug in the values I find out that they are wrong, does it have to do with the fact that I square the equation twice? if so what is the best way to go about solving this equation?

Jean-Claude Arbaut
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Gravity
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  • Line 3 above has an error. You did the equivalent of expanding $(a+2b)^2$ as:$$(a+2b)^2=a^2+4ab+b^2$$while the correct expansion is:$$(a+2b)^2=a^2+4ab+4b^2$$Plus the error spotted below by @Crostul – Mufasa Apr 12 '15 at 12:28
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    There is a mistake in the very first step, when you pass to squares. The square of $2\sqrt{x-5}$ is $4(x-5)$, not $(x+5)$. – Crostul Apr 12 '15 at 12:28
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    Apart from possible mistakes in your computations, bear in mind that you only prove implications ($\Rightarrow$), hence if $x$ is a solution of your equation, then it must be one of the values you find in the end. But that does not mean that all values are solutions. Or even that at least one is a solution, as your equation may also have no solution at all. You have to check afterwards. – Jean-Claude Arbaut Apr 12 '15 at 12:30
  • oh thank you I have spent 1 hour on this dumb mistake:) – Gravity Apr 12 '15 at 12:30
  • well as far as I know if the discriminant is greater than 0 then the equation will have two real solutions, is this wrong? As in no complex numbers will be involved – Gravity Apr 12 '15 at 12:32
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    You original equation is not a quadratic equation. When you square, usually you introduce new "solutions". A trivial example: you square equation $x=1$, then it becomes $x^2=1$, which has solutions $+1$ and $-1$, one of which is of course not a solution of $x=1$. That's the kind of problem that may arise. When you only prove implication in one direction, you have to check implication in the other direction, otherwise it may simply be wrong. – Jean-Claude Arbaut Apr 12 '15 at 12:37
  • By the way, a solution of your equation must be $>5$, otherwise there is a problem with the square root. Here, your second solution is $<5$, so it certainly isn't a solution of your equation. – Jean-Claude Arbaut Apr 12 '15 at 12:40
  • On line 4, it's $3x+7$, not $3x+17$, if I'm not mistaken. – Bernard Apr 12 '15 at 12:41
  • thanks Jean that makes a lot of sense – Gravity Apr 12 '15 at 14:51

2 Answers2

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You forgot a factor $4$ and wrote $x+5$ instead of $x-5$ in the third line, which should be $$ 5x+19=x+7+4\sqrt{x+7}\,\sqrt{x-5}+4(x-5) $$ giving $$ 4\sqrt{x+7}\,\sqrt{x-5}=32 $$ or $$ \sqrt{x+7}\,\sqrt{x-5}=8 $$ that becomes, after squaring, $$ x^2+2x-99=0 $$ The roots of this are $-11$ and $9$, but only the latter is a solution of the original equation, because the existence conditions on the radicals give \begin{cases} 5x+19\ge0\\[3px] x+7\ge0\\[3px] x-5\ge0 \end{cases} that is, $x\ge5$.

egreg
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  • why u expect all the numbers must be positive>????@egreg – David Apr 12 '15 at 13:14
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    @gloom Otherwise the square roots don't exist in the real numbers. Note that the equation is meaningless in the complex numbers, because the square root is not uniquely defined, unless one discusses about branches. – egreg Apr 12 '15 at 13:16
  • oh..you are right..i will make correction in my post... – David Apr 12 '15 at 13:18
  • but... $\sqrt{-1}$ is uniquely defined know???@egreg – David Apr 12 '15 at 13:22
  • @gloom Of course not: which one do you choose between $i$ and $-i$? – egreg Apr 12 '15 at 13:26
  • @gloom $i^2=(-i)^2=-1$ so no. – Karl Apr 12 '15 at 13:27
  • $i$ only know..@egreg – David Apr 12 '15 at 13:28
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    @gloom There is no function $f$ on the complex numbers such that $f(z)^2=z$ for every $z$ and $f(z_1z_2)=f(z_1)f(z_2)$ for all $z_1,z_2$. Note that the latter property is essential for the computation of roots of the original equation. – egreg Apr 12 '15 at 13:29
  • is it not a convention?? – David Apr 12 '15 at 13:29
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    @gloom Writing $i=\sqrt{-1}$ is a bad, but sadly widespread abuse of notation. Carelessly manipulating $\sqrt{x}$ for $x<0$ is not only bad, it's wrong, as explained in previous comments. – Jean-Claude Arbaut Apr 12 '15 at 14:10
  • thanks egreg for your reply, that does illuminate many things, in any case the complex numbers do not even constitute a field – Gravity Apr 12 '15 at 14:53
  • @Gravity Your method was sound, weren't for the mistake in writing $x+5$ instead of $4(x-5)$. Usually I advice working out the existence conditions, so one has not to check the discovered roots back in the original equation. – egreg Apr 12 '15 at 15:00
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    @Gravity "in any case the complex numbers do not even constitute a field". Ahem... Yes, it's a field. But that does only guarantee properties of addition and multiplication, not roots. $\Bbb Q$ is also a field, but as you probably know, $\sqrt{2}$ does not exist in $\Bbb Q$. The real problem with square roots of complex numbers is there are two of them. Thus it's not a function. However, $\sqrt{x}$ is correctly defined as a function, on $\Bbb R^+$. – Jean-Claude Arbaut Apr 12 '15 at 18:23
  • sorry I meant to say, it is not an ordered field because x^2 >= 0 must be the case for ordered fields, so the difference between C, Q, and R is that, Q and R are ordered fields, but C is not, that is what I meant to say. But yeah you probably know that already:=), just correcting my mistake. AGAIN! – Gravity Apr 12 '15 at 20:23
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\begin{align} \sqrt{5x+19}&=\sqrt{x+7}+2\sqrt{x-5}\\ 5x+19&=x+7+4(x-5)+4\sqrt{(x+7)(x-5)}\\ 32&=4\sqrt{(x+7)(x-5)}\\ 8&=\sqrt{(x+7)(x-5)}\\ 64&=x^2+2x-35\\ 0&=x^2+2x-99 \end{align}

which gives $x=9$ and $x=-11$ are the solutions.

But both are giving something like this..

$x=9\implies \sqrt {5(9)+19}=\sqrt{9+7}+2\sqrt{9-5}$ $\implies\sqrt{64}=\sqrt{16}+2\sqrt{4}\implies 8=8$ correct know...!!!!

$x=-11\implies\sqrt {5(-11)+19}=\sqrt{-11+7}+2\sqrt{-11-5}$ $\implies\sqrt{-36}=\sqrt{-4}+2\sqrt{-16}\implies 6i=2i+4i$

But by the series of comments given by the well wishers.. i could understand $x=-11$ is not possible...

David
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