Solving a quadratic is four separate skills that look like one, and a learner who is shaky on any of them loses marks on questions they could otherwise do. This set separates them deliberately.
The first is the habit that has to come before everything else: the equation must equal zero. Two questions here arrive with terms on both sides, and each carries a wrong option built by setting the two sides to zero separately — a step that looks like algebra and is not, because the rule being used applies only to a product that equals zero. Getting this wrong produces an answer that feels earned, which is what makes it expensive.
The second is reading roots correctly out of brackets. A bracket vanishes at the opposite of the number printed inside it, and several wrong options in this set are exactly the numbers as written, signs unreversed. Three questions ask for something slightly different from the roots themselves — the larger of the two, or their sum — because a learner who solves and then stops has answered a question nobody asked. The sum question also points out that the roots of a quadratic with a leading coefficient of one always add to the negative of the middle coefficient, so that answer was available without solving at all.
The third is the formula, for the equations that do not factor. Two questions end in exact expressions containing a square root, and their wrong options are the three things that actually go wrong: dividing only part of the numerator, dividing by the leading coefficient instead of twice it, and losing the sign of the middle coefficient when the formula begins by negating it.
The fourth is counting solutions without finding them. One question asks for the constant that leaves exactly one solution and shows that the result is a perfect square, which is what one solution looks like; another asks only how many real solutions an equation has and is settled by a sign.
Two questions put a quadratic into a situation — a ball thrown from a roof, a garden with a known area — where the algebra offers two roots and one of them is impossible. Solving is not finishing: a negative time is before the throw, and a negative length is not a length. In each, one wrong option is the rejected root and another is the pair handed back undecided.
Every physical quantity is plausible, every place invented, and nothing is drawn from any official publication.
Written for this catalogue, with no source document. The content is the nonlinear-equation material named in the College Board's own public description of the Advanced Math domain — quadratic equations in one variable, the number of solutions, and quadratic models in context; every question, option and explanation here is original, and nothing is reproduced from any official publication or released test. Note that the digital test also asks some questions in a student-produced response format, which a multiple-choice bank cannot reproduce. Zestly is an independent study tool. It is not affiliated with the College Board, which owns the SAT, and it is not an exam centre.
What are the solutions to $x^2 - 7x + 10 = 0$?
$x = 2$ and $x = 5$
The constant is positive and the middle coefficient negative, so both numbers in the brackets are negative: $(x - 2)(x - 5) = 0$. A product is zero only when one of its factors is, so $x = 2$ or $x = 5$, and substituting either back gives zero. The pair $-2$ and $-5$ takes the numbers with the signs printed in the brackets rather than the values that make each bracket vanish, which reverses both. The other two pairs flip exactly one sign, and each of them fails: at $x = -5$ the expression is $70$, and at $x = -2$ it is $28$.