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Solve the equation

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

My reference gives the only solution as $-1$. I can indeed verify this solution but don't have any clue of how to solve for it.

I think squaring might a possible but that seems cumbersome.

David G. Stork
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Sooraj S
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3 Answers3

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Introduce substitution $x^2+2x=y$. Your equation becomes:

$$\sqrt{3y+7}+\sqrt{5y+14}=4-y$$

Now you can do the double squaring and you will end up with equation that is much easier to handle. I hope you can proceed from here.

Saša
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  • The method leads to http://mathworld.wolfram.com/BiquadraticEquation.html and squaring immediately introduces https://math.stackexchange.com/questions/1513842/extraneous-roots which demands re-validation. – lab bhattacharjee Apr 14 '19 at 11:01
3

Observe also that: $$3x^2+6x+7 = 3(x^2+2x+1) + 4 = 3(x+1)^2+4 \ge 4,$$

$$ 5x^2+10x+14 = 5(x^2+2x+1)+9 =5(x+1)^2+9 \ge 9$$

$\implies$ LHS $\ge 2+3 = 5$,

and RHS$ = 5 - (x+1)^2 \le 5$

LHS = RHS needs $(x+1)^2 = 0$

$\implies x=-1$ is the only solution.

lab bhattacharjee
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DeepSea
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1

It's $$\sqrt{3(x^2+2x)+7}+\sqrt{5(x^2+2x)+14}+x^2+2x=4.$$ We see that the expression in the left side increases as a function of $x^2+2x$ and the right side is a constant.

Thus, our equation has roots for one value of $x^2+2x$ maximum.

But $$x^2+2x=-1$$ is valid, which gives the answer:

$$\{-1\}$$

Michael Rozenberg
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