AMC 10 · 2002 · #10
Grade 8 algebraSuppose that a and b are nonzero real numbers, and that the equation x2+ax+b=0 has solutions a and b. Then the pair (a,b) is
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The equation $x^2 + ax + b = 0$ is built from two numbers $a$ and $b$, and its two solutions turn out to be those very same numbers $a$ and $b$. Both $a$ and $b$ are nonzero. Find the pair $(a,b)$ that makes this happen.
Givens: $a$ and $b$ are nonzero real numbers; The quadratic equation is $x^2 + ax + b = 0$; The two solutions of this equation are $a$ and $b$ themselves; Answer choices: (A) $(-2,1)$, (B) $(-1,2)$, (C) $(1,-2)$, (D) $(2,-1)$, (E) $(4,4)$
Unknowns: The values of $a$ and $b$ — that is, the pair $(a,b)$
Understand
Restated: The equation $x^2 + ax + b = 0$ is built from two numbers $a$ and $b$, and its two solutions turn out to be those very same numbers $a$ and $b$. Both $a$ and $b$ are nonzero. Find the pair $(a,b)$ that makes this happen.
Givens: $a$ and $b$ are nonzero real numbers; The quadratic equation is $x^2 + ax + b = 0$; The two solutions of this equation are $a$ and $b$ themselves; Answer choices: (A) $(-2,1)$, (B) $(-1,2)$, (C) $(1,-2)$, (D) $(2,-1)$, (E) $(4,4)$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #3 Eliminate Possibilities, #6 Guess and Check
The whole sentence 'the roots are $a$ and $b$' is a word fact, so Tool #13 (Convert to Algebra) turns it into symbols: rebuild the quadratic in factored form and match it term-by-term to $x^2+ax+b$, which produces two equations linking $a$ and $b$. Tool #3 (Eliminate Possibilities) then uses the 'nonzero' rule to throw away a stray branch when one equation could split two ways. Tool #6 (Guess and Check) confirms the winning pair by plugging it back and reading off the roots.
Execute — Answer: C
6.EE.A.3 Step 1 Rebuild the quadratic from its roots
- A quadratic whose two solutions are $a$ and $b$ can be built straight from those roots: $x^2+ax+b = (x-a)(x-b)$.
- Multiplying the right side out gives $(x-a)(x-b) = x^2 - (a+b)x + ab$.
- Now the exact same quadratic is written in two ways at once.
💡 If you know where a parabola hits zero, you can rebuild its equation by multiplying the two '$x-\text{root}$' pieces.
8.EE.C.8 Step 2 Match the like parts
- Two ways of writing the same quadratic must agree term by term.
- Matching the $x$-coefficients gives $a = -(a+b)$, and matching the constant terms gives $b = ab$.
- So the one sentence about roots has become two equations in $a$ and $b$: $a = -(a+b)$ and $b = ab$.
💡 If two expressions are equal for every $x$, the number in front of each power has to match.
8.EE.C.7 Step 3 Apply the nonzero rule to the constant equation
- Look at $b = ab$.
- Move everything to one side, $ab - b = 0$, then factor: $b(a-1)=0$.
- This could mean $b=0$ or $a=1$.
- But the problem promises $b$ is nonzero, so the $b=0$ branch is ruled out — leaving $a = 1$.
💡 A product is zero only when one factor is zero, and the 'nonzero' promise tells you which factor it must be.
8.EE.C.7 Step 4 Solve for b and read off the pair
- Put $a=1$ into the first equation $a = -(a+b)$: it becomes $1 = -(1+b)$, so $1 = -1 - b$, which gives $b = -2$.
- The pair is $(a,b)=(1,-2)$, which is choice (C).
💡 With one unknown pinned down, the leftover equation hands you the other in a single step.
6.EE.A.3 A quadratic whose two solutions are $a$ and $b$ can be built straight from those 8.EE.C.8 Two ways of writing the same quadratic must agree term by term. Matching the $x$ 8.EE.C.7 Look at $b = ab$. Move everything to one side, $ab - b = 0$, then factor: $b(a-1 8.EE.C.7 Put $a=1$ into the first equation $a = -(a+b)$: it becomes $1 = -(1+b)$, so $1 = Review
Reasonableness: Check the pair by rebuilding the equation: with $a=1$ and $b=-2$ it reads $x^2+x-2=0$, which factors as $(x-1)(x+2)=0$, so its roots are $1$ and $-2$ — exactly $a$ and $b$, and both are nonzero as required. It is worth noting the algebra also permits $a=b=-\tfrac12$, but that pair is not offered among the choices, so among the five options only $(1,-2)$ survives.
Alternative: Skip the factoring and use Vieta's shortcut: for $x^2+ax+b=0$ the two roots add to $-a$ and multiply to $b$. Since the roots are $a$ and $b$, that says $a+b=-a$ (so $2a+b=0$) and $ab=b$ (so $a=1$, because $b\neq0$); then $b=-2$. Or, even faster, just test the five answer choices — only $(1,-2)$ builds an equation, $x^2+x-2=0$, whose roots match its own coefficients.
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(x-a)(x-b)$ into $x^2-(a+b)x+ab$ so the rebuilt quadratic can be lined up against $x^2+ax+b$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Reading the root condition as a system by matching coefficients — the two equations $a=-(a+b)$ and $b=ab$ in the unknowns $a$ and $b$.)8.EE.C.7Solve linear equations in one variable (Solving $b(a-1)=0$ for $a=1$ (using $b\neq0$) and then $1=-(1+b)$ for $b=-2$.)
⭐ If a quadratic's own roots are its coefficients, rebuild it as $(x-a)(x-b)$, match the pieces to $x^2+ax+b$, and let the 'nonzero' rule pick the real answer.
⭐ If a quadratic's own roots are its coefficients, rebuild it as $(x-a)(x-b)$, match the pieces to $x^2+ax+b$, and let the 'nonzero' rule pick the real answer.
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