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Worked solution

Solve a reciprocal equation using the quadratic formula

Find, in terms of kk, the roots of the equation x+1x=kx+\dfrac1x=k.

CJC · O Level revision · Example 2

Answer

x=k±k2−42x=\dfrac{k\pm\sqrt{k^2-4}}2. For real kk, real roots exist exactly when ∣k∣≥2|k|\ge2.

Compare your reasoning

Clear the nonzero denominator

x+1x=k ⟹ x2+1=kx ⟹ x2−kx+1=0\begin{aligned}&x+\frac1x\\ & =k\ \Longrightarrow\ x^2+1\\ & =kx\ \Longrightarrow\ x^2-kx+1\\ & =0\end{aligned}

The original expression requires x≠0x\ne0. Multiplying by x gives the printed quadratic.

Substitute all three coefficients

x=k±(−k)2−4(1)(1)2=k±k2−42\begin{aligned}&x\\ & =\frac{k\pm\sqrt{(-k)^2-4(1)(1)}}2\\ & =\frac{k\pm\sqrt{k^2-4}}2\end{aligned}

As in the source callout, use a=1, b=−k, c=1.

Check the real domain

k2≥4k^2\ge4

Independent qualification of the printed formula: k2−4≥0k^2-4\ge0 means ∣k∣≥2|k|\ge2. The product of the quadratic roots is 1, so neither is zero. For k=2 the repeated root is 1; for k=−2 it is −1.

How did this question go?

Solution notes

The source prints the root formula without stating a real-k restriction. The real-root condition and nonzero check here are independently supplied.