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Find the sum of all the solutions of the equation $|2x^2+x-1|=|x^2+4x+1|$

Though I tried to lớn solve it in desmos.com & getting the requisite answer but while solving it manually it is getting very lengthy.

I tried to construct the two parabola và mirror image the region below y axis but still getting it is getting complicated.

Is there any easy method lớn solve it and get the sum of all the solutions ?


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We are asked for the sum of the roots; we don"t necessarily have khổng lồ find the roots.

Squaring, we get $(2x^2+x-1)^2=(x^2+4x-1)^2$.

So $4x^4+4x^3...=x^4+8x^3...$.

So $colorblue3x^4-colorblue4x^3....=0$.

By Vieta"s formulas, the answer is $dfrac43$.

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The expressions between the absolute value bars have the same or opposite signs. Hence there are two independent cases (by addition & subtraction):

$$3x^2+5x=0$$ & $$x^2-3x-2=0.$$

Then by Vieta,

$$-frac53+3.$$

For complete rigor, one should show that no root is repeated. This is true, because the polynomials have no double root, và their $ extgcd$ is $1$.


$$|2x^2+x-1|=|x^2+4x+1|\(2x^2+x-1)^2=(x^2+4x+1)^2\(2x^2+x-1)^2-(x^2+4x+1)^2=0\<(2x^2+x-1)+(x^2+4x+1)>cdot<(2x^2+x-1)-(x^2+4x+1)>=0\(3x^2+5x)cdot(x^2-3x-2)=0\xcdot(3x+5)cdot(x^2-3x-2)=0\Solvingspace forspace allspace cases,space wespace get:\x=0\x=-frac53\x=frac3pmsqrt172$$


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Squaring and factorizing we get$$x(3x+5)(x^2-3x-2)=0$$The solutions are given by $$x=-frac53lor x=0lor x=frac12 left(3-sqrt17 ight)lor x=frac12 left(3+sqrt17 ight)$$


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