Each of the equations $3x^2 - 2 = 25, (2x - 1)^2 = (x - 1)^2, \sqrt{x^2 - 7} = \sqrt{x - 1}$ has
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two integral roots
- no root greater than $3$
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no root zero
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only one root
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only negative root and one positive root
1) 3x^2 = 27 -> x^2 = 9 -> x = 3, -3. 2) (2x-1)^2 = (x-1)^2 -> 2x-1 = x-1 (x=0) or 2x-1 = -(x-1) (3x=2, x=2/3). 3) sqrt(x^2-7) = sqrt(x-1) -> x^2-7 = x-1 -> x^2-x-6=0 -> (x-3)(x+2)=0. x=3 (valid), x=-2 (invalid). Roots are 3, -3, 0, 2/3, 3. None are greater than 3.
Solving the first equation 3x^2 - 2 = 25 gives 3x^2 = 27 and x^2 = 9, which yields roots of 3 and -3. Solving the second equation (2x - 1)^2 = (x - 1)^2 gives roots of 2 and 0. Solving the third equation sqrt(x^2 - 7) = sqrt(x - 1) gives x^2 - x - 6 = 0, which factors to (x - 3)(x + 2) = 0 and yields roots of 3 and -2. Checking all resulting roots, the highest value among them is exactly 3, meaning none of the equations have any root strictly greater than 3.