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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 ?  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\$.

Xem thêm: Cảm Nhận Của Anh Chị Về Bài Thơ Việt Bắc (Tố Hữu), Top 6 Mẫu Cảm Nhận Bài Thơ Việt Bắc Hay Chọn Lọc  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\$\$ 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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Find the possible values of the parameter \$m\$ for which the given equation has all of its roots real.

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