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The main question is :-

Solve for $x$ :- $$(\sqrt{3x^2+6x+7} + \sqrt{5x^2+10x+4}) = 4-2x-x^2$$

My approach :-

For convenience, I assume $$t_1=3x^2+6x+7$$ $$t_2=5x^2+10x+4$$

Then, by rationalizing, we get,

$$\frac{t_1-t_2}{\sqrt{t_1}-\sqrt{t_2}}=4-2x-x^2$$ Which, when simplified, gives,

$$\frac{2x^2+4x-3}{\sqrt{t_1}-\sqrt{t_2}}=x^2+2x-4$$

I can't go any further. A hint shall be sufficient. Thanks!

EDIT : I have tried squaring and getting rid of the radical root, but trust me it's not worth it IMO. I'd appreciate some other, quicker method if you have any.

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  • $\begingroup$ This should be a trivial problem if you just square the equality, simplify, and then square again to remove the middle term, i.e. $t_1+2\sqrt{t_1t_2}+t_2=(4-2x-2x^2)^2$, simplify with the radical alone on one side, and then square again. $\endgroup$ – JAustin Jul 27 '16 at 13:37
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    $\begingroup$ That leaves me with a 8-power polynomial, which for a young brain like mine is tough. $\endgroup$ – Akshar Gandhi Jul 27 '16 at 13:38
  • $\begingroup$ Exactly, I think it you should try looking that the LHS is always positive but RHS is not $\endgroup$ – Weijie Chen Jul 27 '16 at 13:39
  • $\begingroup$ @Weijie, are you saying that I put the RHS positive and solve for $x$ via inequalities? I guess that would work, but how would you find out other answers if there were any? How will you even say that $RHS>0$ will give us the desired result and not something incomplete? $\endgroup$ – Akshar Gandhi Jul 27 '16 at 13:43
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    $\begingroup$ If it is not a contests problem trust me that it probably don't have a known way to represent the root, so try this with some approaching methods such has Newton's or Horner's $\endgroup$ – Weijie Chen Jul 27 '16 at 14:05
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If the equation to solve is

$$\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+4}=4-2x-x^2$$

then, as others have noted, the main simplification to make is to complete all the squares, rewriting the equation as

$$\sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2-1}=5-(x+1)^2$$

at which point we can let $t=(x+1)^2$ and obtain the simpler-looking equation

$$\sqrt{3t+4}+\sqrt{5t-1}=5-t$$

Before doing any more algebra, note that that any real value of $t$ solving this last equation must satisfy ${1\over5}\le t\le 5$, since the left hand side acquires an imaginary part if $t\lt{1\over5}$, while the two sides have opposite signs if $t\gt5$. Note also that the curve $y=\sqrt{3t+4}+\sqrt{5t-1}$ is strictly increasing for $t\ge{1\over5}$ while the line $y=5-t$ is strictly decreasing, so there can be at most one real solution $t$. Finally, since $$\sqrt{{3\over5}+4}+\sqrt{{5\over5}-1}=\sqrt{{23\over5}}+0\lt\sqrt{45\over5}=3\lt5-{1\over5}$$

while

$$\sqrt{15+4}+\sqrt{25-1}\gt0=5-5$$

we can conclude there is exactly one real value of $t$ that solves the equation. In fact we can see that ${1\over5}\lt t\lt1$ by noting that

$$\sqrt{3+4}+\sqrt{5-1}=\sqrt7+\sqrt4\gt\sqrt4+\sqrt4=4=5-1$$

A brief digression:

If the equation were

$$\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2$$

then the same approach leads to

$$\sqrt{3t+4}+\sqrt{5t+9}=5-t$$

which also features a strictly increasing left hand side and a strictly decreasing right hand side, so can have at most one real solution. But in this case it's easy to see that $t=0$ is a solution, so we conclude that $t=0$ is the only real solution, which gives $x=-1$ as the only real solution of the original (altered) equation.

End of digression.

For the equation $\sqrt{3t+4}+\sqrt{5t-1}=5-t$, unfortunately, there doesn't seem to be any simple solution as in the digression; if you do the algebra of repeated squarings, you get the ugly quartic

$$t^4-36t^3+308t^2-860t+500=0$$

which, according to Wolfram Alpha, doesn't seem to factor nicely. It does, though, have just one real root in the requisite range, $t\approx0.77971$, from which we get $x=-1\pm\sqrt t\approx-1\pm0.883$.

In my opinion, if this was meant to be a preparation problem for a math competition, then there was a typo, changing the number $14$, as in the digression above, into a $4$. The problem as stated can, as I've indicated, be solved numerically, but it's messy and tedious and not the sort of thing one would expect in a competition.

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  • $\begingroup$ Yep, that's the answer my professor gave. Thanks! $\endgroup$ – Akshar Gandhi Jul 28 '16 at 9:02
  • $\begingroup$ @AksharGandhi Did he used the computer ? How he calculated $$t^4-36t^3+308t^2-860t+500=0$$ $\endgroup$ – Aakash Kumar Jul 28 '16 at 9:11
  • $\begingroup$ @AakashKumar, the first squaring, to $3t+4+2\sqrt{(3t+4)(5t-1)}+5t-1=25-10t+t^2$, hardly requires a computer, nor does rewriting it as $2\sqrt{15t^2+17t-4}=t^2-18t+22$. The second squaring, to $60t^2+68t-16=(t^2-18t+22)^2=t^4-36t^3+368t^2-792t+484$, benefits from modern technology in expanding the trinomial, but could be done by hand in a pinch. Finding the roots of the quartic, however, is, these days, best relegated to the computer. $\endgroup$ – Barry Cipra Jul 28 '16 at 12:35
  • $\begingroup$ @BarryCipra Expanding in not a problem , find a root of quartic is a problem specially when roots are not integer , question type this will never come in competitive exam unless some help is provided. $\endgroup$ – Aakash Kumar Jul 28 '16 at 14:57
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Complete the square, as I mentioned in the comments and as Aakash posted in his answer. Then we get: $$ \sqrt{3(x+1)^2 + 4} + \sqrt{5(x+1)^2 - 1} = -(x+1)^2 + 5$$

Now let $t = (x+1)^2$. Then we have:

$$ \sqrt{3t + 4} + \sqrt{5t - 1} = -t + 5$$

Solve this for $t$ the "usual" way: Isolate one radical, square both sides, simplify, isolate the remaining radical term, square both sides, simplify, solve for $t$, verify the solution(s). Then back-substitute to find $x$ and verify those solutions as well.

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  • $\begingroup$ This does not help, the remaining equation is a 4th degree equation which still is not going anywhere. $\endgroup$ – Weijie Chen Jul 27 '16 at 14:20
  • $\begingroup$ I guess that's pretty much what we have to do now... $\endgroup$ – Akshar Gandhi Jul 27 '16 at 14:28

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