Question #44538

a) Solve the equation 9x4 – 18x3 – 31x2 + 8x + 12 = 0 by Ferrari’s method.
b) Solve the equation in (a) above, given that the sum of two of its roots is zero.

Expert's answer

Answer on Question #44538 - Math - Other

a) Solve the equation 9x418x331x2+8x+12=09x^4 - 18x^3 - 31x^2 + 8x + 12 = 0 by Ferrari's method.

We have an equation ax4+bx3+cx2+dx+e=0ax^4 + bx^3 + cx^2 + dx + e = 0

where a=9a = 9, b=18b = -18, c=31c = -31, d=8d = 8, e=12e = 12.


alpha=ca3b28a2=4.944444alpha = \frac{c}{a} - \frac{3b^2}{8a^2} = -4.944444beta=b38a3bc2a2+da=3.555556beta = \frac{b^3}{8a^3} - \frac{bc}{2a^2} + \frac{d}{a} = -3.555556gamma=3b4256a4+cb216a3bd4a2+ea=0.7291667gamma = \frac{-3b^4}{256a^4} + \frac{cb^2}{16a^3} - \frac{bd}{4a^2} + \frac{e}{a} = 0.7291667P=alpha212gamma=2.766461P = \frac{-alpha^2}{12} - gamma = -2.766461Q=alpha3108+alphagamma3beta28=1.662767Q = \frac{-alpha^3}{108} + alpha \frac{gamma}{3} - \frac{beta^2}{8} = -1.662767Rp=Q2+Q24+P327=0.8313837+0.3049106iR_p = \frac{Q}{2} + \sqrt{\frac{Q^2}{4} + \frac{P^3}{27}} = -0.8313837 + 0.3049106iRm=Q2Q24+P327=0.83138370.3049106iR_m = \frac{Q}{2} - \sqrt{\frac{Q^2}{4} + \frac{P^3}{27}} = -0.8313837 - 0.3049106iU=Rm(13)=0.57407410.7698004iU = R_m^{\left(\frac{1}{3}\right)} = 0.5740741 - 0.7698004iy=5alpha6U+P3U=2.972222y = -5 \frac{alpha}{6} - U + \frac{P}{3U} = 2.972222W=alpha+2y=1W = \sqrt{alpha + 2y} = 1


The roots are:


x1=b4a+W3alpha+2y+2betaW2x_1 = \frac{-b}{4a} + \frac{W - \sqrt{3alpha + 2y + 2 \frac{beta}{W}}}{2}x2=b4a+W+3alpha+2y+2betaW2x_2 = \frac{-b}{4a} + \frac{-W + \sqrt{3alpha + 2y + 2 \frac{beta}{W}}}{2}x3=b4a+W3alpha+2y+2betaW2x_3 = \frac{-b}{4a} + \frac{W - \sqrt{3alpha + 2y + 2 \frac{beta}{W}}}{2}x4=b4a+W3alpha+2y+2betaW2x_4 = \frac{-b}{4a} + \frac{-W - \sqrt{3alpha + 2y + 2 \frac{beta}{W}}}{2}x1=3,x2=23,x3=1,x4=23x_1 = 3, x_2 = \frac{2}{3}, x_3 = -1, x_4 = -\frac{2}{3}


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