AMC 10 Step by Step

2013 AMC 10B Problem 19

P16-20~6 minPrint

The real numbers c,b,ac,b,a form an arithmetic sequence with a≥b≥c≥0a\ge b\ge c\ge 0 . The quadratic ax2+bx+cax^2+bx+c has exactly one root. What is this root?

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Key insight

Exactly one root means b^2 = 4ac, and b is halfway between a and c. Take a = 1, solve for c, and the root is -b/(2a).

Solution

1. The numbers c,b,ac, b, a form an arithmetic sequence, so bb is exactly halfway between aa and cc:
b=a+c2.b = \frac{a + c}{2}.
2. A quadratic has exactly one root when the part under the square root in the quadratic formula is zero: b2−4ac=0b^2 - 4ac = 0. Then the formula gives the root as x=−b2ax = -\frac{b}{2a}.
3. The number aa cannot be 00: since a≥b≥c≥0a \ge b \ge c \ge 0, that would make all three 00, and there would be no quadratic. So a>0a > 0.
4. Multiplying aa, bb and cc all by the same positive number does not change the root or either condition. So we may choose a=1a = 1 to keep things simple. Then b=1+c2b = \frac{1 + c}{2}, and b2=4cb^2 = 4c becomes
(1+c)24=4c⟹1+2c+c2=16c⟹c2−14c+1=0.\frac{(1 + c)^2}{4} = 4c \quad\Longrightarrow\quad 1 + 2c + c^2 = 16c \quad\Longrightarrow\quad c^2 - 14c + 1 = 0.
5. By the quadratic formula,
c=14±196−42=14±1922=14±832=7±43.c = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm \sqrt{192}}{2} = \frac{14 \pm 8\sqrt{3}}{2} = 7 \pm 4\sqrt{3}.
6. We need c≤a=1c \le a = 1. Since 7+437 + 4\sqrt{3} is about 13.913.9, it is too big. So c=7−43c = 7 - 4\sqrt{3}, which is about 0.070.07.
7. Then b=1+c2=8−432=4−23b = \frac{1 + c}{2} = \frac{8 - 4\sqrt{3}}{2} = 4 - 2\sqrt{3}.
8. The root is
x=−b2a=−4−232=−2+3.x = -\frac{b}{2a} = -\frac{4 - 2\sqrt{3}}{2} = -2 + \sqrt{3}.

As a sanity check, this is about −0.27-0.27: negative, as it should be when all the coefficients are positive.

The answer is (D) −2+3\boxed{\textbf{(D)}\ -2+\sqrt{3}}.

Why this works

Both conditions stay true when aa, bb, cc are all scaled by the same number, so only their ratios matter, and setting a=1a = 1 leaves just one unknown. The condition a≥ca \ge c is what picks the right sign in step 6. For a quadratic with one root, remember the root is simply −b2a-\frac{b}{2a}.

Alternative approach

Work with the root rr directly. A quadratic with the single root rr and a=1a = 1 is (x−r)2=x2−2rx+r2(x - r)^2 = x^2 - 2rx + r^2, so b=−2rb = -2r and c=r2c = r^2. The arithmetic sequence condition 2b=a+c2b = a + c becomes −4r=1+r2-4r = 1 + r^2, or r2+4r+1=0r^2 + 4r + 1 = 0, so r=−2±3r = -2 \pm \sqrt{3}. We also need b≤ab \le a, that is −2r≤1-2r \le 1, so r≥−12r \ge -\frac{1}{2}. Only r=−2+3≈−0.27r = -2 + \sqrt{3} \approx -0.27 passes.

Concise solution

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

An arithmetic sequence c,b,ac, b, a means bb is the average: b=a+c2b = \frac{a + c}{2}. Exactly one root means the discriminant vanishes: b2=4acb^2 = 4ac. Substituting,
(a+c)24=4ac⟹a2−14ac+c2=0.\frac{(a + c)^2}{4} = 4ac \quad\Longrightarrow\quad a^2 - 14ac + c^2 = 0.
Here a>0a > 0 (otherwise everything is 00 and there is no quadratic), so divide by a2a^2 and set t=cat = \frac{c}{a}:
t2−14t+1=0⟹t=7±43.t^2 - 14t + 1 = 0 \quad\Longrightarrow\quad t = 7 \pm 4\sqrt3.
Since c≤ac \le a, we need t≤1t \le 1, so t=7−43t = 7 - 4\sqrt3.

The double root of ax2+bx+cax^2 + bx + c is x=−b2ax = -\frac{b}{2a}. With b=a+c2b = \frac{a + c}{2},
x=−a+c4a=−1+t4=−8−434=−2+3.x = -\frac{a + c}{4a} = -\frac{1 + t}{4} = -\frac{8 - 4\sqrt3}{4} = -2 + \sqrt3.

The answer is (D) −2+3\boxed{\textbf{(D)}\ -2+\sqrt{3}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.