Solve x² + 1 = 0

x² + 1 = 0 has no real solution, and the reason is short: rearranged it says x² = −1, and no real number squares to a negative. Positive times positive is positive, negative times negative is positive, and zero squares to zero. The list is complete, so no real number is left. The graph agrees. The ...

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.

NO REAL ROOTS

x = i or −i

0 + 1i · 0 − 1i

The discriminant is negative, so no real number satisfies the equation. The two complex roots are exact, and they differ only in the sign of the imaginary part.

As typedx^2 + 1 = 0
Expandedx² + 1 = 0
Standard formx² + 1 = 0

EACH ROOT PUT BACK, EXACTLY

i → 0 + 0i

−i → 0 + 0i

sum 0 against −b/a = 0

product 1 against c/a = 1

DISCRIMINANT

−4

negative, so the roots are complex

AXIS OF SYMMETRY

x = 0

the roots sit either side of this

REAL PART

0

shared by both roots

FACTORS OVER ℚ

no

no real roots, so no real factors

WHAT IS HAPPENING

1Expand both sides and collect

x² + 1 = 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 + 1 = 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)(1) = −4

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.

4A conjugate pair, with an imaginary part

x = i and x = −i

A negative discriminant means no real number satisfies the equation — the curve misses the axis entirely. The two roots differ only in the sign of their imaginary part, which is what makes them a conjugate pair.

THE CURVEthe curve misses the axis entirely, which is why there is no real root−1.51.53.7−0.46the gold circle is the turning point at (0, 1), and it never reaches the axis

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 ÷ 2athe step is imaginary, so neither root lands on this linecentre 0, step ±ia negative discriminant makes the step imaginary — the centre is real, the roots are not

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

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HOW TO USE

  1. 1

    Rearrange to x² = −1 and check whether any real number squares to a negative. None does, so there is no real solution.

  2. 2

    Confirm with the discriminant: 0² − 4(1)(1) = −4. Negative means the curve misses the axis entirely.

  3. 3

    For the complex answer, use i, defined by i² = −1. The roots are x = i and x = −i.

  4. 4

    Check both: i² + 1 = −1 + 1 = 0, and (−i)² + 1 = i² + 1 = 0, since squaring removes the sign here as everywhere.

WORKED EXAMPLE

x² + 1 = 0 Subtract 1 from both sides: x² = −1 No real number satisfies this. A positive number squared is positive, a negative number squared is also positive, and zero squared is zero — that covers every real number, and none of them gives −1. The discriminant confirms it without rearranging: b² − 4ac = 0² − 4(1)(1) = −4, and a negative discriminant means no real roots. On a graph, the curve y = x² + 1 is the ordinary parabola lifted one unit upward. Its lowest point is (0, 1), a full unit above the axis, so it never touches down. In the complex numbers there is an answer. The imaginary unit i is defined so that i² = −1, and then: x = i and x = −i Check: i² + 1 = −1 + 1 = 0. And (−i)² + 1 = (−1)²i² + 1 = i² + 1 = 0. Why bother? Because with complex numbers admitted, every quadratic has exactly two roots and every polynomial of degree n has exactly n — no exceptions, no special cases. The tidiness is not decoration; a great deal of engineering, particularly anything involving oscillation, is built on it.

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

There is no real answer. In the complex numbers the roots are x = i and x = −i, where i is defined by i² = −1. Both check out: i² + 1 = −1 + 1 = 0, and (−i)² gives the same, because squaring removes the sign.

  • ·No real number satisfies it
  • ·x = i and x = −i in the complex numbers
  • ·i is defined by i² = −1
  • ·Both substitute back to zero

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Last updated: August 31, 2026 · Exact surds throughout, never floating point · The equation does not have to be in standard form before you type it.