2002 AMC 10B Problem 10

P1-10~2 minPrint

Quadratics Apply an identity: SFFT, sum of squares, difference of cubes, Vieta

Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2 + ax + b = 0$ has solutions $a$ and $b$ . Then the pair $(a,b)$ is

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Key insight

With solutions a and b, the quadratic is (x-a)(x-b). Matching terms gives a+b = -a and ab = b, so a = 1 and b = -2.

Solution

1. A quadratic that starts with $x^2$ and has solutions $a$ and $b$ can be written as $(x - a)(x - b)$ . Multiply it out:
$(x - a)(x - b) = x^2 - (a + b)x + ab.$
2. This must be the same as $x^2 + ax + b$ . Match the parts:
- The $x$ terms: $-(a + b) = a$ , which means $a + b = -a$ .
- The number terms: $ab = b$ .
3. Start with $ab = b$ . The problem says $b$ is not zero, so we can divide both sides by $b$ and get $a = 1$ .
4. Put $a = 1$ into $a + b = -a$ : $1 + b = -1$ , so $b = -2$ .
5. Check: $x^2 + x - 2 = (x - 1)(x + 2)$ , whose solutions are $1$ and $-2$ . Those are $a$ and $b$ .

The answer is $\boxed{\textbf{(C)}\ (1,-2)}$ .

Why this works

For a quadratic starting with $x^2$ , the solutions add to the negative of the $x$ coefficient and multiply to the constant term. These facts turn "these numbers are the solutions" into two short equations, with no quadratic formula needed. Watch for "nonzero" in a problem: it is usually the permission you need to divide.

Alternative approach

Use the choices. From step 2, $ab = b$ with $b$ not zero forces $a = 1$ . Only choice (C) has $a = 1$ , and it passes the check $a + b = 1 + (-2) = -1 = -a$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

For a monic quadratic $x^2 + ax + b$ , the sum of the roots is $-a$ and the product is $b$ . Here the roots are $a$ and $b$ themselves, so
$a + b = -a, \qquad ab = b.$

Because $b \ne 0$ , dividing the second equation by $b$ gives $a = 1$ . Substituting into the first: $1 + b = -1$ , so $b = -2$ .

Check: $x^2 + x - 2 = (x+2)(x-1)$ has roots $1$ and $-2$ , which are exactly $a$ and $b$ .

The answer is $\boxed{\textbf{(C)}\ (1,-2)}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
Introduction to Algebra, Chapter 11: Special Factorizations
Introduction to Algebra, Chapter 13: Quadratic Equations – Part 2 (§13.2, §13.3)