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

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 ?

• Vieta's formula. Jul 1, 2019 at 13:02
• Note: $0$ is a root Jul 1, 2019 at 13:03
• @EclipseSun We can use Vieta's formula, but to solve a quadratic equation should also not be a real challenge. Jul 1, 2019 at 13:27
• How can a question showing an effort AND context receive $4$ close-votes and $3$ down-votes ? Jul 1, 2019 at 13:30

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 $$\dfrac43$$.

• inspired by Dr. Sonnhard Graubner (squaring) and Eclipse Sun (Vieta's formulas) Jul 1, 2019 at 13:22
• You can even do with $4x^4+4x^3+\cdots=x^4+8x^3+\cdots$.
– user65203
Jul 1, 2019 at 13:52
• Yes @YvesDaoust; I thought of that after I posted Jul 1, 2019 at 13:54

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

• And the conditions for that? Jul 1, 2019 at 13:01
• Since we have $|a|=|b|$ if and only if $a=b$ or $a=-b$ exactly the solutions of those equations solve the given equation. Jul 1, 2019 at 13:04
• Better is, i think, to square both sides of the equation. Jul 1, 2019 at 13:07
• @Dr.SonnhardGraubner This gives an equation of degree $4$. I prefer this approach. Jul 1, 2019 at 13:15
• @YvesDaoust I know because only the sum has to be determined. But my approach is already easy enough and answers the question whether there is an easy solution. Jul 1, 2019 at 13:54

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,

$$-\frac53+3.$$

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 $$1$$.

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

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

• And the conditions for that ?
– user65203
Jul 1, 2019 at 13:06