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SAMATICA
In this chapter: Equations and inequalities
  1. Solve a quadratic equation
  2. Solve cubic, quartic and quintic equations, including complex roots
  3. Solve polynomial inequalities of degree 2, 3 and 4
  4. Solve simultaneous linear equations in 2, 3 or 4 unknowns
  5. SOLVE
  6. Factor, expand and simplify algebraic expressions

Solve polynomial inequalities of degree 2, 3 and 4 in Scientific calculator plus 991 for Android

Press MODE and choose the Solve inequality of degree 2, 3 or 4 entry for your relation, >, ≥, < or ≤ against zero. Type the coefficients, pressing = after each, and press = again. The solution set comes back as intervals, one to a line and read as "or": x² − x − 2 > 0 gives x < −1 on one line and 2 < x on the next.

Modes
degree 2, 3 or 4, one for each of >, ≥, < and ≤ against 0
You type
the coefficients, highest power first
You get
intervals, one per line, read as "or"
Bounds
the real roots; closed for ≥ and ≤, open for > and <

The answer to an inequality is a set of values rather than a number, and the calculator gives it as one: a short list of intervals, with the real roots of the polynomial as their ends. The inequality has to compare a polynomial with 0, so move every term to one side first.

The four relations

Each degree has a separate mode for each relation. They share the same Coefficient Editor and the same steps, and differ only in the comparison with 0.

RelationMeansIts bounds
> 0Strictly greaterOpen, written with <
≥ 0Greater or equalClosed, written with ≤
< 0Strictly lessOpen
≤ 0Less or equalClosed
The inequality solvers in the MODE dialog: each entry shows the inequality it solves, ax²+bx+c > 0 and so on, above the words Solve inequality of degree 2, 3 or 4.
The inequality solvers in the MODE dialog: each entry shows the inequality it solves, ax²+bx+c > 0 and so on, above the words Solve inequality of degree 2, 3 or 4.

Solve it

To solve a polynomial inequality

  1. Press MODE and choose the entry for your degree and relation. Each shows the inequality it solves, such as ax²+bx+c > 0, above the words Solve inequality of degree 2, 3 or 4. The Coefficient Editor appears, one cell for each coefficient.

  2. Type the coefficients from the highest power down, pressing = after each. Enter 0 for a term the polynomial leaves out.

  3. x² − x − 2 > 0 solved: x < −1 on the first line, 2 < x on the second. The letters at the top left only sketch the shape of the answer.

    Press = again, or tap SOLVE [=]. The whole solution set appears on one page.

  4. Press AC or = to go back to the editor and solve again.

Two rays

x² − x − 2 > 0

in
abc
1 -1 -2

out [x < -1; 2 < x]

x² − x − 2 = (x + 1)(x − 2), so the parabola opens upward and is positive outside its roots, −1 and 2.

Reading the answer

The solution set is stacked one interval to a line, and the lines are read as "or": x is a solution when it satisfies any one of them. Each line takes one of four shapes, with the real roots of the polynomial as its ends.

  • x < a: everything to the left of a.
  • a < x: everything to the right of a.
  • a < x < b: everything strictly between two roots.
  • x = a: a single isolated point, at a repeated root where the polynomial only touches 0.

With ≥ and ≤ the ends are included, and the lines use ≤; with > and < they are left out. Complex roots are ignored: they are not points on the number line, so they cannot be ends.

The same coefficients with ≤ instead of > give the complementary set, between the roots and including them:

One interval

x² − x − 2 ≤ 0

out [-1 ≤ x ≤ 2]

A perfect square with ≤ collapses to a single point: (x − 1)² ≤ 0, entered as a = 1, b = −2, c = 1, gives x = 1 alone, on a single line.

Degree 3

A cubic changes sign at each root it crosses, so its solution set alternates from one root to the next.

Three roots, two pieces

x³ − 2x² − x + 2 > 0

in
abcd
1 -2 -1 2

out [-1 < x < 1; 2 < x]

x³ − 2x² − x + 2 = (x + 1)(x − 1)(x − 2), so the sign alternates across −1, 1 and 2.

At a repeated root the curve touches 0 without crossing, and the sign does not change there. With ≥ or ≤ that touching point is part of the answer, either inside a neighbouring interval or on a line of its own:

An isolated point

x³ − 10x² + 33x − 36 ≥ 0

out [x = 3; 4 ≤ x]

Entered as a = 1, b = −10, c = 33, d = −36. The polynomial is (x − 3)²(x − 4): at 3 it only touches 0, and from 4 on it is positive. With > 0 instead, the touching point drops out and 4 < x is left alone.

Degree 4

A quartic works the same way, with up to four real roots for ends.

Two real roots

x⁴ − 3x² − 4 < 0

in
abcde
1 0 -3 0 -4

out [-2 < x < 2]

x⁴ − 3x² − 4 = (x² − 4)(x² + 1). The factor x² + 1 has no real root, so the only ends are −2 and 2.

Three lines

x⁴ − 4x³ − 12x² ≥ 0

out [x ≤ -2; x = 0; 6 ≤ x]

Entered as a = 1, b = −4, c = −12, d = 0, e = 0. The polynomial is x²(x − 6)(x + 2), and the double root at 0 is an isolated point between the two rays.

No solution, or every x

When the curve never crosses the relation, there are no intervals to show, and a dialog says so instead.

  • No solution: the set is empty. x² + x + 1 < 0 has none, because that parabola is positive everywhere; x⁴ + 1 < 0 has none either.
  • Infinite solution: every real number is a solution, as for x² + x + 1 > 0 or x⁴ + 1 > 0.

A cubic always crosses 0, so it always has a solution and is never solved by every x. With every coefficient 0, the inequality becomes 0 compared with 0: No solution for > and <, Infinite solution for ≥ and ≤. OK returns to the editor.

See it on a graph

Press GRAPH at any time, while typing the coefficients or with the answer on screen. Two things are drawn together: the curve of the polynomial with its roots marked, and a shaded region, which is the solution set itself.

The shading is a vertical band running the whole height of the graph, because the condition only involves x. Its width along the x-axis is the answer: read it from left to right. It is not the area between the curve and the axis; the curve is only there to show where the polynomial crosses 0.

x² − x − 2 > 0 drawn: the parabola with its roots −1 and 2 marked, and the two shaded bands, everything left of −1 and everything right of 2.
x² − x − 2 > 0 drawn: the parabola with its roots −1 and 2 marked, and the two shaded bands, everything left of −1 and everything right of 2.

Change the relation and only the shading moves: with < instead of >, the same curve is shaded between −1 and 2.

At a glance

KeyWhat it does
MODEChoose the Solve inequality of degree 2 entry, or 3 or 4, for your relation
=Evaluates a cell and moves on; once every cell is in, solves; with the answer showing, back to the editor
left right Move between the cells
ACEmpties every cell; with the answer showing, back to the editor
GRAPHPlots the curve with its roots, and shades the solution set

Tap the answer to switch it between fractions and decimals; an irrational end such as √2 is then shown either as an exact radical or as a decimal.

Questions

How do I solve a quadratic inequality?

Move everything to one side so the inequality compares a polynomial with 0. Then press MODE, choose the Solve inequality of degree 2 entry for the relation, type a, b and c, and press =. x² − x − 2 > 0 gives x < −1 and 2 < x, on two lines read as "or"; with ≤ instead, the same coefficients give −1 ≤ x ≤ 2.

Why are there two or three lines in the answer?

Because the solution set is not always one interval. Each line is one piece of it, and a value of x is a solution when it satisfies any one line. x⁴ − 4x³ − 12x² ≥ 0 needs three: x ≤ −2, the single point x = 0, and 6 ≤ x.

What does x = 3 mean in an inequality answer?

It is an isolated point: a repeated root where the polynomial touches 0 without changing sign. It only appears with ≥ or ≤, which include the value 0 itself. x³ − 10x² + 33x − 36 ≥ 0, which is (x − 3)²(x − 4), gives x = 3 on one line and 4 ≤ x on the next.

What if every x is a solution, or none is?

Then there are no intervals to show, and a dialog says so instead: No solution when the set is empty, as for x² + x + 1 < 0, and Infinite solution when every real number satisfies the inequality, as for x² + x + 1 > 0.

Related pages

Written from the calculator's own manual and from the app itself, version 7.6.0; every figure is one the manual shows or a capture of the app proves. Last revised on 26 September 2026.

Bug reports and feature requests go to kimcuc@samatica.com — a person reads it.