Solve the equation $|2x^2+x-1|=|x^2+4x+1|$ [closed]

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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 solve it in desmos.com and getting the requisite answer but while solving it manually it is getting very lengthy.

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

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

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5 Answers

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We are asked for the sum of the roots; we don't necessarily have to 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 $\color{blue}3x^4-\color{blue}4x^3....=0$.

By Vieta's formulas, the answer is $\dfrac43$.

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Hint : Solve the two equations $$2x^2+x-1=x^2+4x+1$$ and $$2x^2+x-1=-x^2-4x-1$$

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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 and subtraction):

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

Then by Vieta,

$$-\frac53+3.$$

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For complete rigor, one should show that no root is repeated. This is true, because the polynomials have no double root, and their $\text{gcd}$ is $1$.

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$$ |2x^2+x-1|=|x^2+4x+1|\\ (2x^2+x-1)^2=(x^2+4x+1)^2\\ (2x^2+x-1)^2-(x^2+4x+1)^2=0\\ [(2x^2+x-1)+(x^2+4x+1)]\cdot[(2x^2+x-1)-(x^2+4x+1)]=0\\ (3x^2+5x)\cdot(x^2-3x-2)=0\\ x\cdot(3x+5)\cdot(x^2-3x-2)=0\\ Solving\space for\space all\space cases,\space we\space get:\\ x=0\\ x=-\frac{5}{3}\\ x=\frac{3\pm\sqrt{17}}{2} $$

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Squaring and factorizing we get$$x(3x+5)(x^2-3x-2)=0$$The solutions are given by $$x=-\frac{5}{3}\lor x=0\lor x=\frac{1}{2} \left(3-\sqrt{17}\right)\lor x=\frac{1}{2} \left(3+\sqrt{17}\right)$$

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