PRMO 2015 Question 3

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8. The equations $x^2 - 4x + k = 0$ and $x^2 + kx - 4 = 0$, where $k$ is a real number, have exactly one common root. What is the value of $k$? [PRE-RMO – 2015]

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Solution:
Let the common root be 'b'. First equation has roots a,b and the second one has b,c.
a + b = 4
ab = k
a = 4-b = k/b
=> 4b - b^2 = k => b^2 = 4b - k ____[1]

Similarly,
b + c = -k
bc = -4
c = -k-b = -4/b
=> b^2 + kb = 4
=> b^2 = 4 - kb________[2]

From [1] and [2]
4 - kb = 4b - k => b = 1
=> k = 3
Answer: 3

Solution 2:

Suppose the two quadratics have a common root $\alpha$. Then

$$\begin{cases} \alpha^2 - 4\alpha + k = 0, \\ \alpha^2 + k\alpha - 4 = 0. \end{cases}$$

Subtracting one equation from the other gives

$$(\alpha^2 + k\alpha - 4) \;-\; (\alpha^2 - 4\alpha + k) \;=\; (k + 4)\,\alpha \;-\; (k+4) \;=\; (k+4)(\alpha - 1) = 0.$$

Hence either $k+4=0$ (i.e.\ $k=-4$) or $\alpha=1.$

• If $k = -4,$ then the two quadratics become identical (both are $x^2 -4x -4=0$), so they share both roots, not exactly one.

• If $\alpha=1$, substitute $\alpha=1$ back into either original equation (e.g.\ $1^2 -4\cdot1 + k = 0$) to find $k=3.$

Finally check that for $k=3,$ the two quadratics

$$x^2 - 4x + 3 = 0 \quad \text{and} \quad x^2 + 3x - 4=0$$

share exactly one root ($x=1$). Thus the required value is $\boxed{{\displaystyle 3}}\mathrm{.}$