Section 6.8

Transformation and Solution of Equations

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Problem 2699

The roots of the equation x3+px+q=0x^{3}+px+q=0 are real under the condition 4p3+27q2⩽04p^{3}+27q^{2} \leqslant 0. When are the roots of the equation x3+ax2+bx+c=0x^{3}+ax^{2}+bx+c=0 real?

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Problem 2700

Solve the equation x4+4x3+11x2+14x+12=0x^{4}+4x^{3}+11x^{2}+14x+12=0 by the substitution x=y+σx=y+\sigma with a suitable choice of σ\sigma.

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Problem 2701

Reduce the following equation to one with integer coefficients and leading coefficient equal to unity, applying a substitution of the form y=axy=ax: 5x4−9x3+15x2−12x+18=05x^{4}-9x^{3}+15x^{2}-12x+18=0.

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Problem 2702

Reduce the following equation to one with integer coefficients and leading coefficient equal to unity, applying a substitution of the form y=axy=ax: 900x4−750x3+375x2−13=0900x^{4}-750x^{3}+375x^{2}-13=0.

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Problem 2703

Prove that the equation a0x2n+a1x2n−2+…+an=0a_{0}x^{2n}+a_{1}x^{2n-2}+\ldots +a_{n}=0 has non-real1 roots, if the equation a0x2n−a1x2n−2+…+(−1)nan=0a_{0}x^{2n}-a_{1}x^{2n-2}+\ldots +(-1)^{n}a_{n}=0 has real ones.

Footnotes

  1. Imaginary numbers are complex numbers with nonzero imaginary part (i.e., not real; not to be confused with purely imaginary numbers). ↩

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Problem 2704

Find the equation whose roots equal the squares of the roots of the equation x5+x3+x2+2x+3=0x^{5}+x^{3}+x^{2}+2x+3=0, and prove that the given equation has imaginary1 roots.

Footnotes

  1. Imaginary numbers are complex numbers with nonzero imaginary part (i.e., not real; not to be confused with purely imaginary numbers). ↩

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Problem 2705

In the equation x4−2x3−3x2+4x−1=0x^{4}-2x^{3}-3x^{2}+4x-1=0, one of the roots equals the square of another. Solve the equation.

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Problem 2706

Solve the equation x4−6mx3+2(4m2+n)x2−6mnx+n2=0x^{4}-6mx^{3}+2(4m^{2}+n)x^{2}-6mnx+n^{2}=0.

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Problem 2707

Solve the equation 4(x2−x+1)3−27x2(x−1)2=04(x^{2}-x+1)^{3}-27x^{2}(x-1)^{2}=0.

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Problem 2708

Under what condition do two of the roots of the equation x3+px+q=0x^{3}+px+q=0 differ only in sign?

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Problem 2709

Prove the theorem: the palindromic equation

ax2n+a1x2n−1+…+a1x+a=0 ax^{2n}+a_{1}x^{2n-1}+\ldots +a_{1}x+a=0

is, by the substitution y=(x+1x−1)2y=\left(\dfrac {x+1}{x-1}\right)^{2}, reduced to solving an equation of degree nn and a quadratic equation.

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Problem 2710

Prove the theorem: a palindromic equation with real coefficients x2n+a1x2n−1+…+a1x+1=0x^{2n}+a_{1}x^{2n-1}+\ldots +a_{1}x+1=0, for which ∣an∣⩽2\left|a_{n}\right| \leqslant 2, has at least one pair of imaginary1 roots.

Footnotes

  1. Imaginary numbers are complex numbers with nonzero imaginary part (i.e., not real; not to be confused with purely imaginary numbers). ↩

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Problem 2711

A polynomial with real coefficients

f(x)=xn+a1xn−1+…+an f(x)=x^{n}+a_{1}x^{n-1}+\ldots +a_{n}

cannot, on the interval (−2,2)(-2, 2), remain in absolute value less than 22 throughout.

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Problem 2712

An equation with real coefficients

x2n+1+a1x2n−1+…+anx=m x^{2n+1}+a_{1}x^{2n-1}+\ldots +a_{n}x=m

has at least one root in the interval (−2,2)(-2, 2), if ∣m∣<2\left|m\right| < 2.

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Problem 2713

An equation with real coefficients

x2n+1+a1x2n−1+…+anx=m x^{2n+1}+a_{1}x^{2n-1}+\ldots +a_{n}x=m

has a root in the interval (−2l,2l)(-2l, 2l), where 2l2n+1=∣m∣2l^{2n+1}=\left|m\right|. (Chebyshev.)

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Problem 2714

Transform the following equation by the indicated substitution (Tschirnhaus transformation): x3−x2−4x+4=0x^{3}-x^{2}-4x+4=0; y=x2+x−1y=x^{2}+x-1.

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Problem 2715

Transform the following equation by the indicated substitution (Tschirnhaus transformation): 2x3−5x2+x+2=02x^{3}-5x^{2}+x+2=0; y=2x2+3x−1y=2x^{2}+3x-1.

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Problem 2716

Transform the following equation by the indicated substitution (Tschirnhaus transformation): x3−6x2+15x−14=0x^{3}-6x^{2}+15x-14=0; 2y=x2−5x+82y=x^{2}-5x+8.

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Problem 2717

Transform the following equation by the indicated substitution (Tschirnhaus transformation): x4+2x3−13x2−14x+48=0x^{4}+2x^{3}-13x^{2}-14x+48=0; y=x2+x−7y=x^{2}+x-7.

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Problem 2718

Find the equation for yy whose roots are expressed in terms of the roots of the given equation in the indicated way: x3+2x−1=0x^{3}+2x-1=0; y1=x1x2y_{1}=x_{1}x_{2}, y2=x1x3y_{2}=x_{1}x_{3}, y3=x2x3y_{3}=x_{2}x_{3}.

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Problem 2719

Find the equation for yy whose roots are expressed in terms of the roots of the given equation in the indicated way: x3−x2−3=0x^{3}-x^{2}-3=0; y1=x1x2+x3−x1y_{1}=\dfrac {x_{1}}{x_{2}+x_{3}-x_{1}}, and so on.

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Problem 2720

Find the equation for yy whose roots are expressed in terms of the roots of the given equation in the indicated way: x3+px+q=0x^{3}+px+q=0; y1=x12+x1x2+x22y_{1}=x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}.

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Problem 2721

Find the equation for yy whose roots are expressed in terms of the roots of the given equation in the indicated way: x3+3x2−4x+3=0x^{3}+3x^{2}-4x+3=0; y1=x1x2−x32x1+x2−2x3y_{1}=\dfrac {x_{1}x_{2}-x_{3}^{2}}{x_{1}+x_{2}-2x_{3}}.

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Problem 2722

Find the equation for yy whose roots are expressed in terms of the roots of the given equation in the indicated way: ax3+3a1x2+3a2x+a3=0ax^{3}+3a_{1}x^{2}+3a_{2}x+a_{3}=0; y1=(x1−x2)(x1−x3)y_{1}=(x_{1}-x_{2})(x_{1}-x_{3}).

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Problem 2723

Solve the equation x3−3ax2+(3a2−b2)x−a3+ab2=0x^{3}-3ax^{2}+(3a^{2}-b^{2})x-a^{3}+ab^{2}=0.

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Problem 2724

Solve the equation x3−3(a−b)x2+(3a2−6ab−b2)x−a3+3a2b+ab2−3b3=0x^{3}-3(a-b)x^{2}+(3a^{2}-6ab-b^{2})x-a^{3}+3a^{2}b+ab^{2}-3b^{3}=0.

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Problem 2725

Solve the equation x3−(a+1)2x2+(2a3+a2+2a−1)x−a4+1=0x^{3}-(a+1)^{2}x^{2}+(2a^{3}+a^{2}+2a-1)x-a^{4}+1=0.

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Problem 2726

Solve the equation x3−5x2+3x+6=0x^{3}-5x^{2}+3x+6=0.

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Problem 2727

Solve the equation x4−12ax3+(45a2−b2)x2−2a(29a2−5b2)x+24a2(a2−b2)=0x^{4}-12ax^{3}+(45a^{2}-b^{2})x^{2}-2a(29a^{2}-5b^{2})x+24a^{2}(a^{2}-b^{2})=0.

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Problem 2728

Solve the equation x4−4(a+1)x3+(5a2+14a+4)x2−(2a3+14a2+12a)x+4a3+8a2=0x^{4}-4(a+1)x^{3}+(5a^{2}+14a+4)x^{2}-(2a^{3}+14a^{2}+12a)x+4a^{3}+8a^{2}=0.

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Problem 2729

Solve the equation 2x4+7x3+10x2+11x+6=02x^{4}+7x^{3}+10x^{2}+11x+6=0.

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Problem 2730

Solve the equation x3−6x−9=0x^{3}-6x-9=0 by Cardano's formula and Ferrari's method.

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Remark.
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A general cubic equation y3+Ay2+By+C=0y^{3}+Ay^{2}+By+C=0 is reduced, by the linear substitution y=x−A3y=x-\dfrac {A}{3}, to the equation x3+px+q=0x^{3}+px+q=0. The latter is solved by the substitution x=u+vx=u+v, which introduces two new unknowns: setting u3+v3=−qu^{3}+v^{3}=-q, 3uv+p=03uv+p=0, we obtain Cardano's formula:

x=−q2+q24+p3273+−q2−q24+p3273. x=\sqrt[3]{-\dfrac {q}{2}+\sqrt{\dfrac {q^{2}}{4}+\dfrac {p^{3}}{27}}}+\sqrt[3]{-\dfrac {q}{2}-\sqrt{\dfrac {q^{2}}{4}+\dfrac {p^{3}}{27}}} .

For the cube roots one should take only those values whose product equals −p3-\dfrac {p}{3} (Hudde's method).

A general quartic equation y4+Ay3+By2+Cy+D=0y^{4}+Ay^{3}+By^{2}+Cy+D=0 is reduced, by the substitution y=x−A4y=x-\dfrac {A}{4}, to the equation x4+ax2+bx+c=0x^{4}+ax^{2}+bx+c=0; Ferrari's method solves such an equation by introducing an auxiliary parameter λ\lambda - a root of the cubic resolvent λ3+2aλ2+(a2−4c)λ−b2=0\lambda^{3}+2a\lambda^{2}+(a^{2}-4c)\lambda -b^{2}=0 - and reducing the original equation to two quadratics:

2x2+a+λ=±2λ(x−b2λ). 2x^{2}+a+\lambda =\pm 2\sqrt{\lambda }\left(x-\dfrac {b}{2\lambda }\right) .
Problem 2731

Solve the equation x3+6x−7=0x^{3}+6x-7=0 by Cardano's formula and Ferrari's method (see problem 2730).

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Problem 2732

Solve the equation x3+3x−2=0x^{3}+3x-2=0 by Cardano's formula and Ferrari's method (see problem 2730).

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Problem 2733

Solve the equation x3+3x−4=0x^{3}+3x-4=0 by Cardano's formula and Ferrari's method (see problem 2730).

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Problem 2734

Solve the equation x3−7x−6=0x^{3}-7x-6=0 by Cardano's formula and Ferrari's method (see problem 2730).

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Problem 2735

Solve the equation x4+3x2+2x+3=0x^{4}+3x^{2}+2x+3=0 by Cardano's formula and Ferrari's method (see problem 2730).

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Problem 2736

Solve the equation x4+2x2+4x+8=0x^{4}+2x^{2}+4x+8=0 by Cardano's formula and Ferrari's method (see problem 2730).

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Problem 2737

Solve the equation x4−x3+3x2−5x+2=0x^{4}-x^{3}+3x^{2}-5x+2=0 by Cardano's formula and Ferrari's method (see problem 2730).

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Problem 2738

Solve the equation x4−17x2−20x−6=0x^{4}-17x^{2}-20x-6=0 by Cardano's formula and Ferrari's method (see problem 2730).

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