Porisma

Worked solution

Recover a quartic and reuse its real roots for intersections

A graphing calculator is not to be used in answering this question.

One root of x4+ax3+5x2−x−10=0x^4+ax^3+5x^2-x-10=0, where aa is real, is x=1+2ix=1+2i. Find aa and the other roots. marks

Hence find the xx-coordinates of the intersection point(s) of y=5(x2−2x4)y=5(x^2-2x^4) and y=x3+3x−1y=x^3+3x-1. marks

[2013/MJC/II/1]

CJC · Review problems · Question 5

Answer

a=−3a=-3; other roots 1−2i,−1,21-2i,-1,2. Intersection xx-coordinates: −1,12-1,\tfrac12.

Use the real-coefficient conjugate pair

The root 1−2i1-2i also occurs. The corresponding quadratic is [x−(1+2i)][x−(1−2i)]=x2−2x+5[x-(1+2i)][x-(1-2i)]=x^2-2x+5. Write f(x)=(x2−2x+5)(x2+bx+c)f(x)=(x^2-2x+5)(x^2+bx+c).

How did this part go?

Compare coefficients and list all other roots

Constant terms give 5c=−105c=-10, so c=−2c=-2. Linear terms give 5b−2c=−15b-2c=-1, hence b=−1b=-1. Cubic terms give a=b−2=−3a=b-2=-3. Thus f(x)=(x2−2x+5)(x2−x−2)=(x2−2x+5)(x−2)(x+1)f(x)=(x^2-2x+5)(x^2-x-2)=(x^2-2x+5)(x-2)(x+1). The other roots are 1−2i,−1,21-2i,-1,2.

How did this part go?

Translate intersections into reciprocal real roots

At an intersection, 5(x2−2x4)=x3+3x−15(x^2-2x^4)=x^3+3x-1, or −10x4−x3+5x2−3x+1=0-10x^4-x^3+5x^2-3x+1=0. Zero is not a solution. Divide by x4x^4: 1/x4−3/x3+5/x2−1/x−10=01/x^4-3/x^3+5/x^2-1/x-10=0. Therefore 1/x1/x is a root of ff. A real intersection requires a real xx, so use only 1/x=−11/x=-1 or 22. The required coordinates are x=−1x=-1 and x=1/2x=1/2. The conjugate pair gives non-real reciprocals and no additional real graph intersections.

How did this part go?

Common mistakes

Only real roots of the reciprocal equation produce x-coordinates of real graph intersections. The nonzero check is required before dividing by x⁴.

Solution notes

Source filename says 2026, but its printed header says 2025/JC2; both are retained without silently changing the source identity.

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