1971 AMC 12 Problem 20

Attempt Problem 20 of the 1971 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 1971 AMC 12 solutions, or check the answer key.

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

20.

The sum of the squares of the roots of the equation x2+2hx=3x^2+2hx=3 is 10.10. The absolute value of hh is equal to:
−1-1
12\frac{1}{2}
23\frac{2}{3}
22
None of these
Answer: E
Concepts:Vieta’s Formulasquadraticalgebraic manipulation
Small hint:
Use the sum and product of the two roots
Big hint:
If the roots are r,s,r,s, then r+s=−2hr+s=-2h and rs=−3rs=-3
Solution:
For roots r,s,r,s, Vieta’s formulas give r+s=−2hr+s=-2h and rs=−3.rs=-3. Hence 4h2=(r+s)2=r2+s2+2rs=10−6=4. \begin{aligned} 4h^2=(r+s)^2 &=r^2+s^2+2rs\\ &=10-6=4. \end{aligned} Therefore ∣h∣=1,|h|=1, which is not among choices A-D because choice A is −1.-1.
Therefore, the correct answer is E.
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