2013 AMC 10B Problem 19

Below is the professionally curated solution for Problem 19 of the 2013 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 10B solutions, or check the answer key.

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Concepts:quadraticarithmetic sequence

Difficulty rating: 1790

19.

The real numbers c,b,ac,b,a form an arithmetic sequence with abc0.a \geq b \geq c \geq 0. The quadratic ax2+bx+cax^2+bx+c has exactly one root. What is this root?

743 -7-4\sqrt{3}

23 -2-\sqrt{3}

1 -1

2+3 -2+\sqrt{3}

7+43 -7+4\sqrt{3}

Solution:

Let the common difference be dd, so c=bdc=b-d and a=b+da=b+d.

A quadratic with exactly one real root has discriminant 00, so b24ac=0b^2-4ac=0. Substituting gives b2=4(b+d)(bd)=4b24d2b^2=4(b+d)(b-d)=4b^2-4d^2, hence 4d2=3b24d^2=3b^2.

Since b,d0b,d\ge0, 2d=3b2d=\sqrt3 b. The double root is b2a=b2(b+d)=12+3=2+3\frac{-b}{2a}=\frac{-b}{2(b+d)}=\frac{-1}{2+\sqrt3}=-2+\sqrt3.

Thus, the correct answer is D .

Problem 19 in Other Years