Porisma

Worked solution

Use a purely imaginary root to factor and transform a quartic

The equation z4−4z3+az2−20z+25=0z^4-4z^3+az^2-20z+25=0, a∈Ra\in\mathbb R, has a root kiki with k∈Rk\in\mathbb R.

(i) Explain clearly why more than one value of kk is possible. mark

(ii) Find the possible exact values of kk, show that a=10a=10, and hence find the roots. marks

(iii) Deduce the roots in Cartesian form of (a) w4+4iw3−10w2−20iw+25=0w^4+4iw^3-10w^2-20iw+25=0 and (b) 25v4−20v3+10v2−4v+1=025v^4-20v^3+10v^2-4v+1=0. marks

[2013/DHS/II/4]

CJC · Review problems · Question 12

Answer

k=±5k=\pm\sqrt5, a=10a=10; z-roots ±5 i,2±i\pm\sqrt5\,i,2\pm i. (iii)(a) ±5,1−2i,−1−2i\pm\sqrt5,1-2i,-1-2i; (b) ±i/5,(2±i)/5\pm i/\sqrt5,(2\pm i)/5.

(i) Explain the two signs

All coefficients are real, so a non-real root kiki has conjugate −ki-ki. Zero is not a root because the constant term is 25. Hence k≠0k\ne0 and the two signs give distinct possibilities.

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(ii) Determine k and a

Substituting z=kiz=ki gives (k4−ak2+25)+i(4k3−20k)=0(k^4-ak^2+25)+i(4k^3-20k)=0. The imaginary part gives 4k(k2−5)=04k(k^2-5)=0. Reject k=0k=0; therefore k=±5k=\pm\sqrt5. The real part then gives 25−5a+25=025-5a+25=0, so a=10a=10.

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(ii) Find all four z-roots

The conjugate pair ±5 i\pm\sqrt5\,i supplies z2+5z^2+5. Write P(z)=(z2+5)(z2+bz+c)P(z)=(z^2+5)(z^2+bz+c). Constant terms give c=5c=5 and cubic terms give b=−4b=-4. Thus P(z)=(z2+5)(z2−4z+5)P(z)=(z^2+5)(z^2-4z+5) and the roots are 5 i,−5 i,2+i,2−i\sqrt5\,i,-\sqrt5\,i,2+i,2-i.

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(iii)(a) Rotate the roots

The w-equation is P(iw)=0P(iw)=0, since (iw)4−4(iw)3+10(iw)2−20(iw)+25(iw)^4-4(iw)^3+10(iw)^2-20(iw)+25 has exactly the required coefficients. Therefore z=iwz=iw and w=−izw=-iz. Applying this to the four z-roots gives w=5,−5,1−2i,−1−2iw=\sqrt5,-\sqrt5,1-2i,-1-2i.

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(iii)(b) Source method 1: reciprocal roots

Zero is not a root of the v-equation. Dividing by v4v^4 gives P(1/v)=0P(1/v)=0. Thus v=1/zv=1/z, giving v=±i/5,(2−i)/5,(2+i)/5v=\pm i/\sqrt5,(2-i)/5,(2+i)/5.

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(iii)(b) Source method 2: scaled roots

Multiply the v-equation by 25: 625v4−500v3+250v2−100v+25=0625v^4-500v^3+250v^2-100v+25=0. This is P(5v)=0P(5v)=0, so z=5vz=5v and v=z/5v=z/5. The roots are again ±i/5,(2+i)/5,(2−i)/5\pm i/\sqrt5,(2+i)/5,(2-i)/5. This retains the source’s alternative method with the coefficient consistency explained in the source notes.

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Common mistakes

The new variable is not the old root: apply w=−iz, v=1/z or v=z/5 as appropriate. Check that zero is excluded before dividing.

Solution notes

Source filename says 2026; printed header says 2025/JC2. Both identities are retained; rendered pages determine radicals and modulus bars.

In the alternative scaled solution on printed 16, one line prints −4(5v), but the preceding expanded coefficient is −100v. The matching polynomial term is −20(5v). This typo is explicitly qualified; original native text retained. Radicals √5 are confirmed in the page images.

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