AMC 10 · 2013 · #19
Grade 8 algebraThe real numbers c,b,a form an arithmetic sequence with a≥b≥c≥0. The quadratic ax2+bx+c has exactly one root. What is this root?
Pick an answer.
AMC 10 2013 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: Three real numbers written in the order c, b, a form an arithmetic sequence, and they satisfy a ≥ b ≥ c ≥ 0. The quadratic ax^2 + bx + c has exactly one root. Find that root.
Givens: c, b, a are in arithmetic sequence (equal gaps between consecutive terms); a ≥ b ≥ c ≥ 0; ax^2 + bx + c has exactly one root
Unknowns: The single root of the quadratic ax^2 + bx + c
Understand
Restated: Three real numbers written in the order c, b, a form an arithmetic sequence, and they satisfy a ≥ b ≥ c ≥ 0. The quadratic ax^2 + bx + c has exactly one root. Find that root.
Givens: c, b, a are in arithmetic sequence (equal gaps between consecutive terms); a ≥ b ≥ c ≥ 0; ax^2 + bx + c has exactly one root
Plan
Primary tool: #13 Convert to Algebra
Secondary: #4 Introduce a Variable, #9 Solve an Easier Related Problem, #14 Extreme Principle
Both facts are verbal: 'arithmetic sequence' and 'exactly one root'. Turn each into an equation, then combine them. Scaling by a shrinks three unknowns to two ratios, so the system collapses to one quadratic in a single variable. Finally the ordering a ≥ b ≥ c picks which of the two solutions is legal.
Execute — Answer: D
8.EE.C.7 Step 1 Turn the sequence into an equation
- In an arithmetic sequence the gap between neighbors is constant.
- For c, b, a that means b - c = a - b.
- Rearranging gives 2b = a + c: the middle term is the average of its two neighbors.
💡 In an evenly spaced list, the middle number sits exactly halfway between its neighbors.
8.EE.A.2 Step 2 Write what 'one root' means
- A quadratic has exactly one root only when its graph touches the x-axis at a single point, so it is a perfect square: ax^2 + bx + c = a(x - r)^2.
- Expanding gives ax^2 - 2ar x + ar^2, so matching coefficients forces b = -2ar and c = ar^2.
- From these, the single root is r = -b/(2a) and b^2 = 4ac.
💡 One root means the parabola just grazes the axis, so the quadratic is a square.
8.EE.C.8 Step 3 Scale so a = 1
- The root does not change if we divide the whole quadratic by a (which is nonzero).
- So work with the ratios p = b/a and q = c/a.
- The two conditions become 2p = 1 + q (from 2b = a + c) and p^2 = 4q (from b^2 = 4ac).
- The root we want is r = -b/(2a) = -p/2.
💡 Only the ratios matter for the root, so pin a to 1 and track two numbers instead of three.
8.EE.C.8 Step 4 Combine into one equation
- From 2p = 1 + q solve for q: q = 2p - 1.
- Substitute into p^2 = 4q to erase q.
- This gives p^2 = 4(2p - 1) = 8p - 4, which rearranges to p^2 - 8p + 4 = 0.
💡 Use one condition to eliminate a variable, leaving a single equation to solve.
8.EE.A.2 Step 5 Solve by completing the square
- Complete the square on p^2 - 8p + 4 = 0.
- Move the constant and add 16 to both sides: p^2 - 8p + 16 = 12, so (p - 4)^2 = 12.
- Taking square roots gives p - 4 = ±\sqrt{12} = ±2\sqrt{3}, hence p = 4 ± 2\sqrt{3}.
💡 Rewriting as a square lets you undo it with a single square root.
8.NS.A.2 Step 6 Pick the valid ratio
- The ordering a ≥ b means b/a = p ≤ 1.
- Since 2\sqrt{3} ≈ 3.46, the option 4 + 2\sqrt{3} ≈ 7.46 is far too big, while 4 - 2\sqrt{3} ≈ 0.54 fits.
- Check the rest: q = 2p - 1 = 7 - 4\sqrt{3} ≈ 0.07, which keeps b ≥ c ≥ 0.
- So p = 4 - 2\sqrt{3}.
💡 Only the smaller solution keeps b from exceeding a.
8.EE.C.7 Step 7 Read off the root
- The single root is r = -p/2 = -(4 - 2\sqrt{3})/2 = -2 + \sqrt{3}.
- So the quadratic's one root is -2 + \sqrt{3}, which is answer (D).
💡 The double root of a perfect square is just -b/(2a), here -p/2.
8.EE.C.7 In an arithmetic sequence the gap between neighbors is constant. For c, b, a tha 8.EE.A.2 A quadratic has exactly one root only when its graph touches the x-axis at a sin 8.EE.C.8 The root does not change if we divide the whole quadratic by a (which is nonzero 8.EE.C.8 From 2p = 1 + q solve for q: q = 2p - 1. Substitute into p^2 = 4q to erase q. Th 8.EE.A.2 Complete the square on p^2 - 8p + 4 = 0. Move the constant and add 16 to both si 8.NS.A.2 The ordering a ≥ b means b/a = p ≤ 1. Since 2\sqrt{3} ≈ 3.46, the option 4 + 2\s 8.EE.C.7 The single root is r = -p/2 = -(4 - 2\sqrt{3})/2 = -2 + \sqrt{3}. So the quadrat Review
Reasonableness: Take a = 1, so p = 4 - 2\sqrt{3} gives b = 4 - 2\sqrt{3} ≈ 0.54 and c = q = 7 - 4\sqrt{3} ≈ 0.07. Then 1 ≥ 0.54 ≥ 0.07 ≥ 0 holds, and c, b, a = 0.07, 0.54, 1 are evenly spaced (gap ≈ 0.46). The discriminant b^2 - 4ac = (4 - 2\sqrt{3})^2 - 4(7 - 4\sqrt{3}) = (28 - 16\sqrt{3}) - (28 - 16\sqrt{3}) = 0, confirming exactly one root at -b/(2a) = -2 + \sqrt{3} ≈ -0.27.
Alternative: Instead of completing the square, feed p^2 - 8p + 4 = 0 into the quadratic formula: p = (8 ± \sqrt{64 - 16})/2 = 4 ± 2\sqrt{3}, then pick p ≤ 1 and compute -p/2 as before. Or test the choices directly: only a root strictly between -1 and 0 is consistent with a positive b/a less than 1, which rules out every option except -2 + \sqrt{3}.
CCSS standards used (min grade 8)
8.EE.C.7Solve linear equations in one variable (Rearranging the arithmetic-sequence condition into 2b = a + c and computing the root -p/2.)8.EE.A.2Use square root symbols to represent solutions and evaluate square roots (Expressing the one-root condition as a perfect square and taking the square root when completing the square.)8.EE.C.8Analyze and solve pairs of simultaneous equations (Scaling by a to two ratios and substituting one equation into the other to get p^2 - 8p + 4 = 0.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Estimating 4 ± 2\sqrt{3} to decide which root satisfies a ≥ b ≥ c ≥ 0.)
⭐ Turn each word-fact into an equation, combine them into one quadratic, then let the size rule a ≥ b ≥ c choose the answer that fits.
⭐ Turn each word-fact into an equation, combine them into one quadratic, then let the size rule a ≥ b ≥ c choose the answer that fits.
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