Overview
Quadratic questions can ask for much more than two numerical roots. The coefficients encode a root sum and product, known roots determine factors, and an equation involving a square root can produce a quadratic whose solutions must be checked against a stronger original condition. This lesson develops the relationships between those forms so that you choose the shortest sound route rather than applying one formula to every problem.
Find a root expression without finding each root
Let r and s be the roots of 2x² − 7x − 4 = 0. Their sum is 7/2 and their product is −2. To find 1/r + 1/s, combine the fractions first: the expression equals (r + s)/(rs). Substituting the known sum and product gives (7/2)/(−2) = −7/4.
The nonzero constant term ensures neither root is zero, so the reciprocals are defined. The alternative 7/4 loses the negative sign of the product; −2 reports the product itself. This is a useful example of working backward from the target: its denominator tells you to inspect the root product, while its numerator becomes the root sum. Solving for both roots would work, but it introduces more arithmetic without adding information needed for the requested expression.
Build the polynomial from its zero factors
A monic quadratic with roots −3 and 5 is (x + 3)(x − 5). Each factor vanishes at the corresponding root, and multiplying the factors gives leading coefficient 1, as “monic” requires. At x = 2, the factors are 5 and −3, so the polynomial value is −15.
The signs in the factors are worth checking deliberately. A root r corresponds to x − r, so a negative root creates a plus sign. The answer 15 drops the sign of the second factor after correct substitution. You do not need to expand the polynomial unless expansion serves the target. Keeping the factored form makes the sign and the zeros more visible. If the polynomial were not specified as monic, its roots alone would leave an unknown multiplier; the leading-coefficient information is therefore part of what makes this question determinate.
Record the sign condition before squaring
For √(2x + 3) = x − 3, the right side must be nonnegative because a principal square root is nonnegative. Thus x ≥ 3. Squaring both sides gives 2x + 3 = x² − 6x + 9, or x² − 8x + 6 = 0. Solving the quadratic yields 4 − √10 and 4 + √10.
Both candidates are positive, so a superficial “discard the negative root” check would not help. The original restriction is stronger: the candidate must be at least 3. Since 4 − √10 is less than 3, it gives a negative right side and cannot equal the principal square root. The larger candidate meets the sign condition and satisfies the squared equality, so it also satisfies the original equality. The unique solution is 4 + √10. This domain check is the main reasoning demand, not simply executing the quadratic formula.
The smaller candidate remains positive because √10 is less than 4, but it is below 3 because √10 is greater than 1. These exact comparisons establish the domain failure without a decimal approximation. They illustrate a useful habit with radical choices: compare a positive radical with an integer by comparing their squares, then translate that bound into the expression being tested.
Distinguish a necessary transformed equation from the original
Squaring preserves every original solution, but it can add new ones because opposite numbers have the same square. That is why solving the squared equation provides candidates rather than a final answer. Substitution or an equivalent sign-and-domain argument must decide which candidates survive. The smaller root in this example is a plausible distractor precisely because it passes the quadratic calculation while failing the original expression.
For a fresh quadratic problem, identify what you actually need: roots, a symmetric root expression, a value of the polynomial, or candidates for another equation. Use coefficient relations for symmetric expressions, factors for known zeros, and complete-square or formula methods when the roots themselves matter. Carry denominator and square-root restrictions alongside the algebra. Before selecting an answer, state which facts guarantee that the result solves the question as written, including its leading coefficient, domain and requested sign or representation.
Apply the method to a new case
Solve √(x + 5) = x − 1 over the reals. Because a square root is nonnegative, require x ≥ 1 before squaring. Squaring gives x + 5 = (x − 1)², or x² − 3x − 4 = 0. The factors are (x − 4)(x + 1), producing candidates 4 and −1. Only 4 survives the original domain, and substitution confirms √9 = 3 = 4 − 1. At −1, the two sides would be 2 and −2, so squaring has erased a sign difference. The final answer is x = 4. Rejecting −1 only because it looks unusual would not teach a reusable check; reject it because it violates the necessary sign condition and the original equation. Every squaring step creates candidates that still require verification.