SAT Mathematics Practice Test 6

The number of solutions to any quadratic equation in the form $\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}=0,$ where a, b, and c are constants, can be found by evaluating the expression ${\mathit{b}}^2-4\mathit{a}\mathit{c},$ which is called the discriminant. If the value of ${\mathit{b}}^2-4\mathit{a}\mathit{c}$ is a positive number, then there will be exactly two real solutions to the equation. If the value of ${\mathit{b}}^2-4\mathit{a}\mathit{c}$ is zero, then there will be exactly onereal solution to the equation. Finally, if the value of ${\mathit{b}}^2-4\mathit{a}\mathit{c}$ is negative, then there will be noreal solutions to the equation.
The given equation $2{\mathit{x}}^2-4\mathit{x}=\mathit{t}$ is a quadratic equation in one variable, where $\mathit{t}$ is a constant. Subtracting t from both sides of the equation gives $2{\mathit{x}}^2-4\mathit{x}-\mathit{t}=0.$ In this form, $\mathit{a}=2,\mathit{b}=-4,$ and $\mathit{c}=-\mathit{t}.$ The values of t for which the equation has no real solutions are the same values of $\mathit{t}$ for which the discriminant of this equation is a negative value. The discriminant is equal to ${\left(-4\right)}^2-4\left(2\right)\left(-\mathit{t}\right);$ therefore, Simplifying the left side of the inequality gives Subtracting 16 from both sides of the inequality and then dividing both sides by 8 gives Of the values given in the options, −3 is the only value that is less than −2. Therefore, choice A must be the correct answer.
Choices B, C, and D are incorrect and may result from a misconception about how to use the discriminant to determine the number of solutions of a quadratic equation in one variable.