Question

Each of the equations $3x^2-2=25,$ $(2x-1)^2=(x-1)^2,$ $\sqrt{x^2-7}=\sqrt{x-1}$ has:

two integral roots

no root greater than $3$

no root zero

only one root

one negative root and one positive root

Answer: B

Concepts:equation solving radical domain common property

Difficulty rating: 1670

Small Hint:

Solve each equation separately and compare the properties of their roots

Big Hint:

For the radical equation, reject candidates that make either radicand negative

Solution:

The first equation gives $x=\pm3.$ Factoring the difference of squares in the second gives $\begin{aligned} &[(2x-1)-(x-1)]\\ &\quad\cdot[(2x-1)+(x-1)]\\ &\qquad=x(3x-2)=0, \end{aligned}$ so $x=0$ or $\frac{2}{3}.$ Squaring the third gives $x^2-x-6=0,$ with candidates $3$ and $-2;$ only $3$ satisfies the real-domain restrictions. Every root obtained is at most $3.$

Thus, each equation has no root greater than $3,$ and the correct answer is B.

1955 AMC 12 Problem 30

Answer: B

Difficulty rating: 1670

Small Hint:

Big Hint:

Solution:

Problem 30 in Other Years