Solve x² − 2 = 0

x² − 2 = 0 gives x = ±√2, and this is the equation that historically broke a whole school of mathematics. The Pythagoreans held that every quantity was a ratio of whole numbers, and the diagonal of a unit square — which is exactly this root — is not. The proof is short. Suppose √2 were a fraction i...

Type the equation as it is written — it does not have to be in standard form. (x − 2)(x + 3) = 5x − 1 is an ordinary quadratic, and rearranging it into ax² + bx + c = 0 is most of the work. Roots come out exactly: fractions where they are rational, simplified surds like (3 + √17)/4 where they are not, and exact complex pairs where the curve misses the axis. If the squared terms cancel, the equation was linear all along and this solves it rather than complaining.

THE EQUATION

Powers with ^, brackets, fractions and terms on both sides all work. Any single letter can be the unknown.

TWO REAL ROOTS

x = √2 or −√2

1.414213… · −1.414213…

Both roots are irrational. The surds above are exact; the decimals below them are rounded.

As typedx^2 - 2 = 0
Expandedx² − 2 = 0
Standard formx² − 2 = 0

EACH ROOT PUT BACK, EXACTLY

√2 → 0 + 0√2

−√2 → 0 + 0√2

sum 0 against −b/a = 0

product −2 against c/a = −2

DISCRIMINANT

8

positive, so two real roots

AXIS OF SYMMETRY

x = 0

the roots sit either side of this

AS DECIMALS

1.4142…, −1.4142…

rounded; the surds above are exact

FACTORS OVER ℚ

no

the roots are irrational

WHAT IS HAPPENING

1Expand both sides and collect

x² − 2 = 0

Brackets multiplied out and powers expanded, with each side written as a multiple of the square, a multiple of the unknown, and a constant.

2Move everything to one side

x² − 2 = 0

Subtracting the right from the left. Only now is the equation in the form the quadratic formula expects, and for most equations as they are actually written this step is the bulk of the work.

3Work out b² − 4ac

(0)² − 4(1)(−2) = 8

This single number decides the shape of the answer: positive gives two roots, zero gives one, negative gives a conjugate pair with an imaginary part.

4Two irrational roots

x = √2 and x = −√2

The discriminant is not a square number, so the roots are irrational. The surd is the exact answer and the decimal beside it is a rounding of it.

THE CURVEthe curve crosses the axis twice, and the crossings are the roots−3.13.19−3.4filled circles are the roots · the gold circle is the turning point at (0, −2)

Why the formula is worth trusting over factoring. Factoring is faster when it works, and it works only when the discriminant is a square number — which across quadratics with two real roots and small coefficients is 21.2% of them. On the other four fifths, time spent hunting for factors is time spent looking for something that is not there. The formula never fails to apply, and this page shows the factorisation whenever one genuinely exists so you can see which case you are in rather than guessing.

HOW THE ROOTS ARE BUILTthe formula gives a centre and a step: the roots sit the same distance either side of the axis0−b ÷ 2a√2−√2centre 0, step ±√2the step is √(b² − 4ac) ÷ 2a, which is zero exactly when the discriminant is

Exact surds throughout · each root is substituted back symbolically, not numerically

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HOW TO USE

  1. 1

    Add 2 to both sides to get x² = 2, then take the square root of both sides — remembering that both a positive and a negative value square to 2.

  2. 2

    Write the answer as x = ±√2. This is exact; any decimal you write instead is not.

  3. 3

    Confirm it does not factor: the discriminant is 0² − 4(1)(−2) = 8, which is not a square number, so no integer factorisation exists.

  4. 4

    Check by substituting: (√2)² − 2 = 2 − 2 = 0, and (−√2)² − 2 = 2 − 2 = 0.

WORKED EXAMPLE

x² − 2 = 0 Add 2 to both sides and take the square root, keeping both signs: x² = 2 x = ±√2 That is the exact answer. As a decimal √2 is about 1.41421356, and no finite decimal is exactly right — 1.414² is 1.999396, and 1.4142² is 1.99996164. Each extra digit gets closer and none arrives. The reason is a proof rather than a limitation of effort. Suppose √2 could be written as a fraction p/q in lowest terms. Then 2 = p²/q², so p² = 2q². That makes p² even, and a number whose square is even must itself be even, so write p = 2k. Substituting gives 4k² = 2q², so q² = 2k², making q even as well. But p and q cannot both be even in a fraction already in lowest terms. The assumption fails, so no such fraction exists. Check both roots: (√2)² − 2 = 2 − 2 = 0, and (−√2)² − 2 = 2 − 2 = 0, because squaring removes the sign. Worth noting that this equation does not factor over the integers. The discriminant is 8, which is not a square number, so no amount of searching will produce whole-number factors — the test tells you that before the searching starts.

REFERENCE RULES

RuleWhat it saysWhy
Standard formax² + bx + c = 0Everything has to reach this shape first. For most typed equations that is the work.
The formulax = (−b ± √(b² − 4ac)) ÷ 2aAlways available. Factoring is quicker when it is possible, which is not often.
The discriminantb² − 4acPositive gives two roots, zero gives one, negative gives a complex pair.
Factoring is the exception21.2% of real-root casesHunting for factors on the rest is time spent on something that is not there.
When it factorsthe discriminant is a square numberThat is the exact test. No guessing required.
The axis of symmetryx = −b ÷ 2aThe roots sit an equal distance either side of it.
Roots as surds(3 + √17)/4, not 1.7808The surd is exact; a decimal put back into the equation misses zero.
The x² term may cancel(x+1)² = x² + 5Then the equation was linear all along, with one solution.
Complex roots come in pairsp + qi and p − qiThey differ only in the sign of the imaginary part.
Sum and product−b/a and c/aA check that never touches the formula that produced the roots.
The formula can cancelwhen b² is far larger than 4acOne root becomes a difference of nearly equal numbers. 90% failed in doubles.
Not linear, not quadraticx³ is refusedA cubic needs different methods entirely, so it gets a reason rather than a number.
The unknown in a denominator1/x is refusedNot a polynomial equation. Multiplying up first may make it one.
Where the formula comes fromcompleting the square, in generalThat method has its own page here; the formula is it done once with letters instead of numbers.

FREQUENTLY ASKED QUESTIONS

x = √2 and x = −√2. Both are exact and neither can be written as a finite decimal. Substituting confirms them: (√2)² − 2 = 0, and (−√2)² gives the same because squaring removes the sign.

  • ·x = √2 and x = −√2
  • ·Both are exact answers
  • ·Neither is a finite decimal
  • ·Both substitute back to zero

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Last updated: September 1, 2026 · Exact surds throughout, never floating point · The equation does not have to be in standard form before you type it.