Problem A Day Marathon – Algebra

  1. Find solutions to $x + 5 + \frac 6 x = 0$.
    Solution Let $x$ be a non-zero real number. Suppose $x + 5 + \frac 6 x = 0$. Let's multiply both sides by $x$: $x^2 + 5x + 6 = 0$.
    Notice that the left-hand side equals $(x+2)(x+3)$.
    Therefore, $x = -2$ or $x = -3$ and, reciprocally, we can check that they're indeed solutions of the equation.
  2. $a$, $b$, $c$ and $d$ are nonnegative real numbers. If we have: $a^2 + b^2 = 1$, $c^2 + d^2 = 1$ and $ac - bd = 12$, find $ad + bc$.
    Solution (1) $a^2c^2 + b^2d^2 - abcd = (ac - bd)^2 = \frac 1 4$
    (2) $a^2d^2 + b^2c^2 + abcd = (ad + bc)^2$
    Therefore, adding (1) and (2): $a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2 = (ad + bc)^2 + \frac 1 4$
    Notice that $a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2 = (a^2 + b^2)(c^2 + d^2) = 1 \cdot 1 = 1$.
    Consequently, $(ad + bc)^2 + \frac 1 4 = 1$, so $(ad + bc)^2 = \frac 3 4$
    Finally, since $ad + bc \ge 0$, we have $ad + bc = \frac {\sqrt 3} 2.
  3. Two numbers sum to 1337 and have a product 9001. Find the two numbers.
    Solution Let $a$ and $b$ be the two numbers, with $a \le b$.
    Then we know that $a$ and $b$ are the solutions of the equation "$x^2 - (a+b)x + abx = 0$", with $a+b = 1337$ and $ab = 9001$. Using the quadratic formula, we find that
    $a = \frac {1337 - \sqrt{1751565}} {2}$ and
    $b = \frac {1337 + \sqrt{1751565}} {2}$
  4. Find the sum of all the real and nonreal roots of $x^{2001} + \left(\frac 1 2 - x\right)^{2001}$
    Solution Binomial expansion: $\left(\frac 1 2 - x\right)^{2001} = \sum_{k=0}^{2001} \left({2001} \choose {k} \right) \frac {1} {2^{2001-k}} (-1)^k x^k$
    Here, the coefficient before $x^{2001}$ equals $-1$, so that term will cancel with $x^{2001)$: the degree of this polynomial is actually 2000.
    Using Vieta's formula: the sum of the roots ($s$) of this polynomial equals $-\frac{a_{1999}} {a_{2000}}$ if we note $a_0, \cdots, a_{2000}$ the coefficients of this polynomial.
    $a_{1999} = \left({2001} \choose {1999} \right) \frac {1} {2^{2001-1999}} (-1)^{1999} = -250 \cdot 2001$
    $a_{2000} = \left({2001} \choose {2000} \right) \frac {1} {2^{2001-2000}} (-1)^{2000} = \frac {2001} {2}$
    Therefore, $s = 500$.
  5. Express $(0^3 - 350)(1^3 - 349)(2^3 - 348) \cdots (350^3 - 0)$ as concisely as possible.
    Solution Well, $7^3 = 343$ so the product is equal to $0$...
    You can also notice it because the product is $\prod_{k = 0}^{350} (k^3 + k - 350)$ and $7$ is a root of the polynomial $x^3 + x - 350$.
    For $k = 7$, we have $k^3 + k - 350 = 0$.
  6. If $a + \frac 1 a = 3$, find $a^4 + \frac {1} {a^4}$.
    Solution $\left(a + \frac 1 a \right)^2 = a^2 + \frac {1} {a^2} + 2 = 3^2 = 9$ so $a^2 + \frac {1} {a^2} = 7$.
    $\left(a^2 + \frac {1} {a^2} \right)^2 = a^4 + \frac {1} {a^4} + 2 = 7^2 = 49$ so $a^4 + \frac {1} {a^4} = 47$.
  7. Let $r$ and $s$ be roots of $x^2 - 5x + 2$. Find $\frac {r^3 - 1} {r - 1} + \frac {s^3 - 1} {s - 1}$.
    Solution According to Vieta's formulas, $r + s = -5$ and $rs = 2$.
    Notice that $r \neq 1$ and $s \neq 1$ (because $1$ isn't a root of this polynomial), so the expression is well defined.
    Moreover, $\frac {r^3 - 1} {r - 1} + \frac {s^3 - 1} {s - 1} = r^2 + r + 1 + s^2 + s + 1$ because of the identity "$a^3 - b^3 = (a-b)(a^2+ab+b^2)$"
    Therefore, $\frac {r^3 - 1} {r - 1} + \frac {s^3 - 1} {s - 1} = (r^2 + s^2) + (r + s) + 2$.
    Notice that $r^2 + s^2 = (r + s)^2 - 2rs = 25 - 4 = 21$.
    Therefore, $\frac {r^3 - 1} {r - 1} + \frac {s^3 - 1} {s - 1} = 21 - 5 + 2 = 18$.