Solve the Following Quadratic Equation for All Values of x in Simplest Form: Complete Method

The equation 2(x² − 9) + 1 = 1 has exactly two solutions: x = 3 and x = −3. Expand the left side to 2x² − 17, then set it equal to 1 to get 2x² = 18, so x² = 9. Taking the square root of both sides yields x = ±3. Both values check out when substituted back into the original equation.

The expression in the prompt's tail, "2 x 2 −9 1 1," looks like a reformatted copy of the equation rather than a second problem. Read literally, the given quadratic is 2(x² − 9) + 1 = 1. That is the equation solved here, step by step, with the reasoning that lets you handle any near neighbor of it.

Why the constant terms cancel cleanly

Quadratics get simpler when you track constants separately from the x-terms. Here the left side builds from 2(x² − 9), which distributes to 2x² − 18, plus 1 gives 2x² − 17. The right side is 1. Subtracting 1 from both sides leaves 2x² − 18 = 0.

The mechanism is cancellation, not magic. Two constants, −1 and +1, remove each other, and what remains is 2x² = 18. The coefficient 2 divides out, giving x² = 9. A square equal to a positive number always has two real roots, one positive and one negative, because squaring erases sign. That is why the answer is a pair, not a single number.

This is the stretched-vertex view: y = 2(x² − 9) + 1 is the parabola y = 2x² shifted down by 17. Its vertex sits at (0, −17), so the curve crosses the line y = 1 exactly twice, at x = −3 and x = 3. Every step above mirrors that picture.

Reading the equation as 2x² − 9 + 1 = 1 changes the answer. That version gives 2x² = 9, so x = ±(3√2)/2 ≈ ±2.121. Twenty-one hundredths apart in position looks small in algebra and is large in application. Where a term sits decides the answer.

The numbers behind the two roots

The discriminant, b² − 4ac, measures how many real roots a quadratic has. In standard form 2x² − 18 = 0, a = 2, b = 0, and c = −18. So the discriminant is 0 − 4(2)(−18) = 144, a perfect square of 12. A positive perfect square means two rational roots that resolve to integers: x = (±12)/4 = ±3. The perfect square is the deeper reason the answer comes out clean.

Verification is cheap and worth doing. At x = 3: 2(9 − 9) + 1 = 1. At x = −3: 2(9 − 9) + 1 = 1. Both hold. The vertex value, −17, sits 17 units below the right side's constant, which is why no root can be fractional in the neighborhood of zero.

A common slip: dropping the coefficient 2. If you divide only x² − 9 = 0, you get x = ±3 for the wrong reason and cannot tell which step mattered. Keep the coefficient attached until division.

Edge cases that change the answer

Several variants of this equation look similar and behave differently.

Two x-terms, 2x² − 9x + 1 = 1: subtract 1, factor out x, get x(2x − 9) = 0, so x = 0 or x = 4.5. A sign flip inside the parentheses, 2(x² + 9) + 1 = 1: gives x² = −9 and only complex roots, x = ±3i. A coefficient on the right, 2(x² − 9) + 1 = 5: gives 2x² − 18 = 4, so x² = 11 and x = ±√11 ≈ ±3.317. A fractional coefficient, (1/2)(x² − 9) + 1 = 1: gives x² = 9 again, since the coefficient cancels at division.

Each case reduces to the shape ax² + c = k. Compare x² to (k − c)/a and check the sign of the right side. Positive means two real roots, zero means one repeated root, negative means a complex conjugate pair.

Misconceptions that cost points

The biggest error is forgetting the negative root. A square equation has two branches, root plus and root minus; dropping one throws away half the solution set. On a placement test or a graded problem, a half solution is a wrong answer.

Equally common is reading √9 as ±3 in the middle of a calculation. The radical symbol by itself means the principal, nonnegative root. The ± sign enters when you take a square root of both sides, which is a different operation than evaluating a radical. Keeping those two ideas separate prevents sign errors.

A third: assuming every quadratic has a clean integer answer. The discrminant rule tells you when it does; 144 here is why the answers land on 3. Most quadratics do not, and answers like ±√11 are correct simplest form. A fourth: simplifying (3√2)/2 by "canceling" a 3. There is nothing to cancel; that number is already in lowest terms.

Pro tip: check whether the factored form matches. 2x² − 18 = 2(x² − 9) = 2(x − 3)(x + 3). Setting each factor to zero gives x = 3 and x = −3. Factoring confirms the algebra independently, and two independent methods agreeing is strong evidence you have not made a sign error.

How to solve this and its neighbors reliably

Expand every bracket, then move all terms to one side in the form ax² + bx + c = 0. Here that yields 2x² − 18 = 0. Divide both sides by the coefficient of x² when the other terms divide evenly. Two constants become 1 and 18. Take the square root of both sides with an explicit ± sign. Lower-school mistakes mostly live in this step. Verify by substitution into the original equation, not the simplified one. That catches errors introduced during simplification. Report both roots in simplest form and separate them with a comma or ±. For anything with an x-term, switch to factoring or the quadratic formula; the square-root method only fits equations with no middle term.

If you run this checklist and get x² equal to a negative number, stop and note complex roots, since a high-school or standardized setting often asks for real solutions only.

When this method does not apply

The square-root method works only when the equation has no x-term, meaning b = 0. Once a middle term appears, factoring or the quadratic formula is the reliable path; the discriminant still tells you how many real roots exist but not where they are. Complex roots are legitimate answers mathematically, though many course levels ask for real solutions only.

Rounding is another limit. If a calculator returns 3.0000001 for a root, do not write decimals when integer roots exist; squared exact values show the answer is 3, not an approximation. And a "simplest form" answer is defined by the course, not by taste. Integer and radical forms are safest where no decimal is requested.

Comparison of methods for x² = 9 in practice

Practice tracking: how skill builds

The time column is a planning guide, not a measurement. Speed comes from the checklist, not from rushing the ± step.

Practical actions that lock in the skill

Write the original equation at the top of a page before touching it. Rewrite it once, unchanged, and only then start moving terms. Ten seconds of verification after simplification catches most sign errors. Keep a personal error log with three columns: step, mistake, fix. After five near-identical equations, most people stop dropping signs.

For exam prep, solve each quadratic two ways. If both methods give the same pair of roots, move on. If they disagree, recheck the factoring, since that step is usually where the error sits. Check whether a radical can be simplified: √9 = 3, √8 = 2√2. In simplest form, a root like ±√11 stays as is, while ±√18 becomes ±3√2.

Finally, translate structure instead of memorizing symbols. A quadratic with no middle term reduces to x² = positive number, and the ± sign takes it from there. The ability to classify the equation before solving it is the real skill.

Limits of what this method settles

This method answers "what are the roots of 2(x² − 9) + 1 = 1" exactly. It does not tell you whether the equation describes a measurement that can be negative, or whether a course wants only real solutions. Those are context questions, not algebra questions. Where the problem asks for all complex values of x, the method extends cleanly: x² = −9 gives x = ±3i using the same two-step logic.

One caveat from test design: equations mined from copied text, like the trailing "2 x 2 −9 1 1," often encode formatting loss. If a second equation was intended, re-copy it exactly before solving, because a single shifted exponent or minus sign changes the answer. The safest habit is to state the intended equation in standard form before solving.