Question

the values $-1$ and $1$ make the equation $3x^2 + 7 = 10$ true. use set notation to express the solution to the equation. (1 point)

$\bigcirc\\ \\{x = -1, x = 1\\}$

$\bigcirc\\ (x = -1, x = 1)$

$\bigcirc\\ (-1, 1)$

$\bigcirc\\ \\{-1, 1\\}$

Explanation:

To express the solution set of an equation, we use set notation where the elements are the solutions. The solutions to the equation \(3x^{2}+7 = 10\) are \(x=- 1\) and \(x = 1\). In set notation, we list the elements (the solutions) inside curly braces \(\{\}\). The notation \(\{x=-1,x = 1\}\) is incorrect because we just list the values, not the equations. The notation \((x=-1,x = 1)\) is incorrect as parentheses are not used for solution sets (parentheses are used for intervals or ordered pairs in some cases). The notation \((-1,1)\) is an interval (representing all real numbers between - 1 and 1), not a set of solutions. The correct set notation for the solutions \(-1\) and \(1\) is \(\{-1,1\}\).

Answer:

\(\{-1,1\}\)