AIME 2007 I · 第 8 题

Solution

Solution 1

We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic.

Let this root be $a$.

We then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k$ , so $Q_{2}(a)-2Q_1(a)=15a+3k=0\implies a = \frac{-k}{5}$.

We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so we get: $\frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k$ or $k^{2}=30k$, so $k=30$ is the highest.

We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $\boxed{030}$.

Solution 2

Again, let the common root be $a$; let the other two roots be $m$ and $n$. We can write that $(x - a)(x - m) = x^2 + (k - 29)x - k$ and that $2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)$.

Therefore, we can write four equations (and we have four variables), $a + m = 29 - k$, $a + n = \frac{43}{2} - k$, $am = -k$, and $an = \frac{k}{2}$.

The first two equations show that $m - n = 29 - \frac{43}{2} = \frac{15}{2}$. The last two equations show that $\frac{m}{n} = -2$. Solving these show that $m = 5$ and that $n = -\frac{5}{2}$. Substituting back into the equations, we eventually find that $k = \boxed{030}$.

Solution 3

Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$, which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$, where $R(x) = ax + b$ and $S(x) = cx + d$. Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$. Equating coefficients, we get the following system of equations:

$$\begin{aligned} a = 2c \\ b = -d \\ 2c(k - 29) - d = c(2k - 43) + 2d \\ -d(k - 29) - 2ck = d(2k - 43) + ck \end{aligned}$$ Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$, we see that the $k$'s drop out and we're left with $d = -5c$. Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\boxed {030}$.

~ anellipticcurveoverq

Solution 4

Notice that if the roots of $Q_1(x)$ and $Q_2(x)$ are all distinct, then $P(x)$ would have four distinct roots, which is a contradiction since it's cubic.

Thus, $Q_1(x)$ and $Q_2(x)$ must share a root. Let this common value be $r.$

Thus, we see that we have

$$r^2 + (k - 29)r - k = 0,$$ $$2r^2 + (2k - 43)r + k = 0.$$ Adding the two equations gives us

$$3r^2 + (3k - 72)r = 0 \implies r = 0, 24 - k.$$ Now, we have two cases to consider. If $r = 0,$ then we have that $Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.$ On the other hand, if $r = 24 - k,$ we see that

$$Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.$$ This can easily be checked to see that it does indeed work, and we're done!

~Ilikeapos

Solution 5

Since $Q_1(x) = x^2 + (k - 29)x - k$ and $Q_2(x) = 2x^2 + (2k - 43)x + k$ are both factors of the cubic polynomial $P(x)$, and both are quadratics, the only way for them to divide the same cubic is if they share a common linear factor. Otherwise, their product would be degree 4, which is too large. Therefore, $P(x)$ can be factored as $P(x) = Q_1(x)(2x + a) = Q_2(x)(x + b)$ for some constants $a, b$. We choose the linear factors so that both products have degree 3 with leading coefficient 2: since $Q_1$ has leading coefficient 1, its linear factor must have leading coefficient 2, and since $Q_2$ has leading coefficient 2, its linear factor must have leading coefficient 1.

Looking at constant terms on both sides, $-k \cdot a = k \cdot b$, so $-k a = k b \implies a = -b.$

Next, expanding both sides and comparing the $x^2$ coefficients gives $a + 2k - 58 = 2b + 2k - 43.$ Substituting $a = -b$, we get $-b + 2k - 58 = 2b + 2k - 43 \implies -b = 2b + 15 \implies b = -5, \quad a = 5.$

Finally, comparing the $x$ coefficients: $a(k - 29) - 2k = (2k - 43) b + k,$ which becomes $5(k - 29) - 2k = (2k - 43)(-5) + k,$ simplifying to $3k - 145 = -9k + 215.$ Solving for $k$ yields $12k = 360 \implies \boxed{30}.$

~MathKing555

Video Solution

https://www.youtube.com/watch?v=bsRQZwO7n84&t=64s