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Quadratic Inequalities: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra I. You can skip it.

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Problem 1 of 10

Problem 1 A number line without its symbol

The number line in the figure is the solution of $(x+2)(x-1)\mathbin{\square}0$ , where the square is one of $<$ , $>$ , $\le$ , or $\ge$ . Find the missing symbol.

The number line given in the problem, from $-4$ to $3$ .

Text description of this figure

A single number line with an arrowhead at each end. It carries a tick and a label at every integer from negative 4 through 3. A filled circle sits at negative 2 and a second filled circle sits at 1, and the stretch of the line between those two circles is shaded. The parts of the line to the left of negative 2 and to the right of 1 carry no shading, and no other point is marked.
1. Hint 1
  The shaded interval tells you which sign of the product is required.
2. Hint 2
  The endpoint markers tell you whether zero is allowed.
Answer

$\le$ .

Full solution

Between $-2$ and $1$ , one factor is positive and the other negative, so the product is negative.
For example, at $x=0$ ,
$(0+2)(0-1)=-2$

The filled endpoints include the two roots, where the product is zero.
The selected values therefore make the product negative or zero, requiring $\le$ .

Answer

$\le$ .

Key idea

A shaded interval and its endpoint markers determine the inequality symbol once the sign of the leading coefficient is known.
Problem 1: Mark complete
Problem 2 A sign record

A quadratic has distinct real roots $-4$ and $2$ , and its value at $x=0$ is negative. Determine whether its value at $x=3$ is positive or negative.
1. Hint 1
  The two roots divide the number line into three sign intervals.
2. Hint 2
  The known value lies between the roots, while the requested input lies to their right.
Answer

Positive.

Full solution

Write the quadratic as $a(x+4)(x-2)$ , with $a\ne0$ .
Its value at $x=0$ is
$a(0+4)(0-2)=-8a$

This is negative, so $a>0$ .
Its value at $x=3$ is
$a(3+4)(3-2)=7a$

That is positive, as expected for the interval to the right of both roots.

Answer

Positive.

Key idea

Knowing the sign in one interval determines the other intervals of a quadratic with two distinct roots.
Problem 2: Mark complete
Problem 3 A solution with a check

Solve $2x^2+9x>5$ . Then confirm the result by testing one value your answer includes.
1. Hint 1
  One side is not yet zero, so gather every term on the left before looking for roots.
2. Hint 2
  The quadratic on the left factors with a leading coefficient of $2$ ; find its two roots.
3. Hint 3
  For an upward parabola, above the axis means the two intervals beyond the roots, and a strict symbol leaves the endpoints out.
Answer

$x<-5$ or $x>\tfrac12$ ; the test value $x=1$ gives $11>5$ , which is true.

Full solution

Subtract $5$ from both sides so that one side is zero.
$2x^2+9x-5>0$

Factor the left side.
$(2x-1)(x+5)>0$

The roots are $x=\tfrac12$ and $x=-5$ .
The parabola opens upward, so the expression is positive outside the roots, and the strict symbol leaves them out.
The solution is $x<-5$ or
$x>\tfrac12.$

Test $x=1$ , which lies inside the right-hand piece.
The left side of the original inequality is $11$ , and $11>5$ is true.
Testing $x=0$ , which lies between the roots, gives $0$ on the left, and $0>5$ is false.

Answer

$x<-5$ or $x>\tfrac12$ ; the test value $x=1$ gives $11>5$ , which is true.

Key idea

Moving every term to one side turns an inequality into a sign question, and a test value confirms the choice of intervals.
Problem 3: Mark complete
Problem 4 Comparing two expressions

Two expressions are $2x^2-x+3$ and $x^2+4x+9$ . Find all real $x$ for which the first is at most the second, and show the solution on the blank number line in the figure.

A blank number line from $-3$ to $8$ , ticked at every integer.

Text description of this figure

A single number line with an arrowhead at each end. It carries a tick and a label at every integer from negative 3 through 8. The line carries no circles, no shading, and no other marks.
1. Hint 1
  Compare the difference of the two expressions with zero.
2. Hint 2
  Find its roots and test one point in each interval.
3. Hint 3
  The inclusive comparison determines whether the endpoints are included.
Answer

$-1\le x\le6$ ; on the number line, filled circles at $-1$ and $6$ with the segment between them shaded.

Full solution

Subtract the second expression from both sides.
$x^2-5x-6\le0$

Its factorization is
$(x+1)(x-6)\le0$

Test $x=-2$ , $0$ , and $7$ in the three intervals.
The products are $8$ , $-6$ , and $8$ , respectively, so the middle interval is negative and the outer intervals positive.
Include roots $-1$ and $6$ because equality is allowed.
On the number line, fill these endpoints and shade between them.

Answer

$-1\le x\le6$ ; on the number line, filled circles at $-1$ and $6$ with the segment between them shaded.

Key idea

Subtracting one expression from another turns a comparison into the sign of a single quadratic.
Problem 4: Mark complete
Problem 5 A moving point

A circle has equation $(x-1)^2+(y+2)^2=25$ . A point has coordinates $(t,1)$ , with real $t$ . Find all values of $t$ for which the point is strictly outside the circle.
1. Hint 1
  Outside means the squared distance from the center is greater than the squared radius.
2. Hint 2
  Substitute the fixed y-coordinate, then find the roots that separate the permitted horizontal positions.
Answer

$t<-3$ or $t>5$ .

Full solution

The center is $(1,-2)$ .
The outside condition is
$(t-1)^2+9>25$

Subtract $16$ after collecting constants.
$(t-1)^2-16>0$

This factors as $(t+3)(t-5)>0$ , with roots $-3$ and $5$ .
The upward quadratic is positive outside these roots.
The roots themselves put the point on the circle, so exclude them.

Answer

$t<-3$ or $t>5$ .

Key idea

A point constrained to a horizontal line can be tested against a circle using a quadratic inequality.
Problem 5: Mark complete
Problem 6 A limit on diagonals

A convex polygon with $n$ sides has $\tfrac{n(n-3)}{2}$ diagonals, where $n$ is a whole number with $n\ge3$ . Find every $n$ for which the polygon has at most $20$ diagonals.
1. Hint 1
  Compare the diagonal count with $20$ , clear the fraction, then move every term to one side.
2. Hint 2
  Factor the resulting quadratic and keep the interval where it is negative or zero.
3. Hint 3
  A polygon needs at least three sides, so discard every value below $3$ .
Answer

$n=3,4,5,6,7,$ or $8$ .

Full solution

The condition on the number of diagonals is
$\frac{n(n-3)}{2}\le20$

Multiply both sides by the positive number $2$ , then move every term to the left.
$n^2-3n-40\le0$

Factor.
$(n-8)(n+5)\le0$

The upward quadratic is zero or negative between its roots, giving
$-5\le n\le8.$
A polygon has at least $3$ sides, so the permitted whole numbers are $3$ through $8$ .
A polygon with $8$ sides has exactly $20$ diagonals, while one with $9$ sides has $27$ , above the limit.

Answer

$n=3,4,5,6,7,$ or $8$ .

Key idea

An inequality's algebraic solution must be cut back to the whole numbers the situation allows.
Problem 6: Mark complete
Problem 7 An exact boundary

Solve $x(2x+3)>4$ . Give exact endpoints.
1. Hint 1
  Expand the product first, then gather every term on one side.
2. Hint 2
  The quadratic does not factor over the whole numbers, so the quadratic formula supplies its roots.
3. Hint 3
  An upward parabola is positive outside its roots, and the strict symbol excludes them.
Answer

$x<\tfrac{-3-\sqrt{41}}4$ or $x>\tfrac{-3+\sqrt{41}}4$ .

Full solution

Expand the left side and move every term to the left.
$2x^2+3x-4>0$

The discriminant is $9+32=41$ , so there are two real roots.
$x=\frac{-3\pm\sqrt{41}}4$

The parabola opens upward, so the expression is positive outside the roots, and the strict symbol excludes them.
Since $\sqrt{41}\approx6.40$ , the roots are about $-2.35$ and $0.85$ .

A check confirms the choice of intervals.
At $x=0$ , which lies between the roots, the left side of the original inequality is $0$ and the right side is $4$ , so that middle interval is correctly left out.

Answer

$x<\tfrac{-3-\sqrt{41}}4$ or $x>\tfrac{-3+\sqrt{41}}4$ .

Key idea

When a rearranged quadratic does not factor, the quadratic formula gives the exact endpoints of the solution.
Problem 7: Mark complete
Problem 8 A divided product

A student solves $(x-1)(x+4)>0$ by dividing by $x-1$ and reporting $x>-4$ . Decide whether this solution is correct, and give the complete solution set.
1. Hint 1
  The divisor can be positive, negative, or zero.
2. Hint 2
  Find the roots of the original product and test the sign in each interval.
Answer

Incorrect; $x<-4$ or $x>1$ .

Full solution

The roots are $-4$ and $1$ .
At $x=-5$ the product is $6>0$ , at $x=0$ it is $-4<0$ , and at $x=2$ it is $6>0$ .
The correct solution is the two outer intervals, excluding the roots.

Division by $x-1$ is invalid at $x=1$ and reverses the inequality when $x<1$ .
Ignoring those conditions loses valid inputs such as $-5$ and includes invalid ones such as $0$ .

Answer

Incorrect; $x<-4$ or $x>1$ .

Key idea

Dividing an inequality by a variable expression requires knowing both its nonzero status and its sign.
Problem 8: Mark complete
Problem 9 A claim with an added constant

For real $k\ge0$ , a student says that $x^2-2x+2+k>0$ holds for every real $x$ . Is the claim correct? Justify it with the discriminant or an equivalent squared form.
1. Hint 1
  A quadratic that never reaches the axis keeps one sign for every input, so ask whether this one can have a real root.
2. Hint 2
  Compare the discriminant with zero, or separate a real square from a positive constant.
Answer

Yes, the inequality holds for every real $x$ and every $k\ge0$ .

Full solution

The discriminant is
$D=4-4(2+k)$

Thus $D=-4-4k<0$ when $k\ge0$ .
The upward quadratic has no real roots and remains positive everywhere.
Its squared form confirms the sign:
$x^2-2x+2+k=(x-1)^2+1+k$

The square is zero or positive and $1+k$ is positive, so their sum is positive for every real input.

Answer

Yes, the inequality holds for every real $x$ and every $k\ge0$ .

Key idea

Adding a zero or positive constant to a strictly positive quadratic preserves its positive sign.
Problem 9: Mark complete
Problem 10 Two possible instructions

Give an example of a quadratic inequality with a negative leading coefficient and no real solutions. Then change only its inequality symbol to make every real number a solution. Justify both results.
1. Hint 1
  Choose a downward quadratic that never reaches the x-axis.
2. Hint 2
  A square with a negative coefficient and a negative constant can have a fixed sign for all real inputs.
Answer

For example, $-x^2-1\ge0$ has no real solutions, while $-x^2-1<0$ holds for all real $x$ .

Full solution

Choose the quadratic $-x^2-1$ .
Its discriminant is
$D=0-4(-1)(-1)$

This is $-4$ , so the quadratic has no real roots.
Since it opens downward, it is negative for every real input.
Therefore $-x^2-1\ge0$ has no solutions, and changing just the symbol to $-x^2-1<0$ admits every real number.

Answer

For example, $-x^2-1\ge0$ has no real solutions, while $-x^2-1<0$ holds for all real $x$ .

Key idea

A quadratic that stays strictly negative yields either no solutions or all real solutions according to the comparison symbol.
Problem 10: Mark complete

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