1964 AMC 12 Problem 30

Attempt Problem 30 of the 1964 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1964 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

30.

If
(7+43)x2+(2+3)x−2=0, \begin{aligned} (7+4\sqrt3)x^2 &+(2+\sqrt3)x\\ &-2=0, \end{aligned}
the larger root minus the smaller root is:
−2+33-2+3\sqrt3
2−32-\sqrt3
6+336+3\sqrt3
6−336-3\sqrt3
33+23\sqrt3+2
Answer: D
Concepts:quadraticradicalalgebraic manipulation
Small hint:
Notice that 7+43=(2+3)27+4\sqrt3=(2+\sqrt3)^2
Big hint:
For Ax2+Bx+C=0,Ax^2+Bx+C=0, the root difference is B2−4ACA\frac{\sqrt{B^2-4AC}}{A}
Solution:
Let A=7+43=(2+3)2A=7+4\sqrt3=(2+\sqrt3)^2 and B=2+3.B=2+\sqrt3. The discriminant is B2−4A(−2)=A+8A=9A.B^2-4A(-2)=A+8A=9A. Hence the root difference is 9AA=32+3=6−33.\frac{\sqrt{9A}}A=\frac3{2+\sqrt3}=6-3\sqrt3.
Therefore, the correct answer is D.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.