Algebra Practice

Inequalities with Parameters Practice Test

Analyze how a parameter changes inequality direction, boundaries, and the number or type of solutions.

Inequalities with Parameters Practice Test

20 parameter-based inequality questions with instant feedback and worked explanations.

Instant feedback · Worked explanations
Inequality Branch Canvas

Before dividing, decide whether the parameter expression is positive, negative, or zero.

Parameterized inequalities require a case split before the usual isolation steps. A positive divisor preserves the inequality direction, a negative divisor reverses it, and a zero divisor cannot be canceled at all. The same case logic extends to target boundaries, absolute-value inequalities, and quadratic inequalities where the discriminant and leading coefficient control whether the graph lies above or below the horizontal axis.

Positive caseDivision keeps the inequality symbol pointing the same way.
Negative caseDivision reverses the inequality symbol.
Zero caseDo not divide. Substitute the parameter value and classify the numerical statement.

1. Split by sign before dividing

A single parameterized inequality can produce three different solution behaviors.

Sign-sensitive inequality
(k−2)x>6
The sign of the coefficient multiplying the variable determines both legality of division and direction of the result.

Positive coefficient

k>2
x>6k−2

The inequality direction stays unchanged.

Negative coefficient

k<2
x<6k−2

The inequality direction reverses after division.

Zero coefficient

k=2
0>6

The resulting false statement means there is no solution.

2. The zero case can mean no solution or all real values

Strict and non-strict inequalities behave differently when the variable term disappears.

Strict inequality at the zero parameter value

(k−3)x<k−3
k=3
0<0

A false strict numerical statement gives no solution.

Non-strict inequality at the same zero value

(k−3)x≤k−3
k=3
0≤0
x∈R

A true statement independent of the variable is satisfied by every real value.

3. Match a requested boundary by replacing the inequality with equality

The equality identifies the candidate boundary; then the sign of the coefficient confirms the required direction.

Requested solution boundary

(k+1)x≥8
x≥2

The boundary occurs where the two sides are equal.

boundary

Recover the parameter

(k+1)·2=8
k=3
k+1>0

The candidate value also makes the variable coefficient positive, so the requested direction is preserved.

4. Absolute-value inequalities change type when the parameter crosses zero

The right-hand parameter controls whether an interval exists, collapses to one point, or becomes impossible.

Strict absolute-value inequality

|x−2|<k
k>0
2−k<x<2+k
k≤0
∅

A strict absolute-value inequality requires a strictly positive radius.

Non-strict absolute-value inequality

|x−2|≤k
k>0
2−k≤x≤2+k
k=0
x=2
k<0

At zero radius, the non-strict interval collapses to exactly one point; negative radius is impossible.

5. Strict and non-strict quadratic inequalities use different discriminant endpoints

For an upward-opening quadratic to stay above the horizontal axis for all real inputs, the repeated-root endpoint matters.

Base quadratic
x2−4x+k
D=16−4k
The leading coefficient is positive, so the discriminant determines whether the graph crosses, touches, or stays above the axis.

Strictly positive for all real inputs

x2−4x+k>0
D<0
k>4

The repeated-root case is excluded because the expression equals zero at one point.

Nonnegative for all real inputs

x2−4x+k≥0
D≤0
k≥4

The repeated-root endpoint is allowed because equality is permitted.

6. For quadratic inequalities, check the leading coefficient and discriminant together

The discriminant alone does not tell whether a quadratic is positive or negative everywhere.

Graph-position filter

To stay positive for all real inputs, the parabola must open upward and avoid the horizontal axis. For a non-strict condition, touching the axis is allowed.

Strict positivity
kx2+2x+k>0
k>0
D=4−4k2
D<0
k>1
Nonnegative version
kx2+2x+k≥0
D≤0
k≥1

The boundary parameter is included only in the non-strict version.

7. When an upward quadratic is negative, the solution lies between its real roots

The parameter must first create two real roots; only then does the open interval between them exist.

Feasibility and boundaries

x2−4x+k<0
k<4
x=2−4−k
x=2+4−k

The strict inequality excludes both root boundaries.

Final interval

2−4−k<x<2+4−k

The upward-opening parabola lies below the horizontal axis only between the two roots.

8. A requested quadratic boundary can also be found by equality

If a particular input must be an endpoint of the solution set, substitute that input and set the quadratic equal to zero.

Target endpoint

x2−6x+k≤0
x=3

A boundary point of a polynomial inequality is normally a root of the corresponding equation.

root

Solve the boundary equation

32−6·3+k=0
k=9

After finding the parameter, verify whether the requested strict or non-strict inequality includes the endpoint.

9. Error analysis: the sign and zero cases must be settled before division

Most parameter-inequality mistakes come from applying a legal step on the wrong branch.

Dividing without checking the sign

x>6k−2 is incomplete because the divisor may be positive, negative, or zero.

Forgetting to reverse the inequality

(k−2)<0 signals the branch where division must reverse the inequality direction.

Canceling the zero case

If the parameter coefficient can equal zero, substitute that value instead of canceling it.

Strict quadratic endpoint included

D=0 is not enough for a quadratic that must be strictly positive everywhere.

Non-strict quadratic endpoint excluded

D≤0 correctly allows the repeated-root case.

Boundary matched but direction not checked

Equality finds the candidate endpoint; the coefficient sign still determines which side of the endpoint is included.

Final parameter-inequality audit

Before accepting a solution set, verify the sign branch, zero case, boundary direction, strictness, and quadratic feasibility.

1
What is the sign of the parameter expression before division?Positive preserves direction, negative reverses it, and zero requires a separate case.
2
If the variable coefficient becomes zero, what numerical statement remains?That decides between all real values and no solution.
3
Was a target boundary given?Use equality to find the candidate, then verify the inequality direction.
4
Does an absolute-value radius cross zero?That can change an interval into one point or no solution.
5
For a quadratic, what is the leading-coefficient sign?Combine graph orientation with the discriminant condition.
6
Is the inequality strict or non-strict?Repeated-root endpoints are excluded from strict conditions but may be included in non-strict ones.
This block supports the Inequalities with Parameters Practice Test. Its examples are illustrative rather than copies of the test questions. The focus is sign-based parameter splitting, zero-coefficient cases, target solution boundaries, absolute-value conditions, all-real and no-solution cases, quadratic inequalities, leading-coefficient and discriminant checks, intervals between roots, and strict versus non-strict endpoints.