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,


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.

  • $\begingroup$ 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

If the equation to solve is


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


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


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 $$\sqrt{{3\over5}+4}+\sqrt{{5\over5}-1}=\sqrt{{23\over5}}+0\lt\sqrt{45\over5}=3\lt5-{1\over5}$$



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


A brief digression:

If the equation were


then the same approach leads to


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


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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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