Core practice ← Back to lesson

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: Core (core-course level)

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 A squared threshold

Solve $2(x-1)^2+3<11$ , giving the answer in interval notation.
1. Hint 1
  Isolate the square to see how far the input may be from its center.
2. Hint 2
  The strict symbol excludes the two boundary distances.
Answer

$(-1,3)$ .

Full solution

$\begin{aligned} (x-1)^2&<4 \\ -2&<x-1 \\ x-1&<2 \end{aligned}$

A real number's square is less than $4$ exactly when the number lies strictly between $-2$ and $2$ .
Adding $1$ throughout translates this to $-1<x<3$ .
At either endpoint the original left side is $11$ , which is excluded.

Answer

$(-1,3)$ .

Key idea

Requiring a square to be less than a positive bound keeps its base strictly between the negative and positive square roots of that bound.
Problem 1: Mark complete
Problem 2 Comparing two rules

Find all real $x$ for which $f(x)=x^2+2x$ is at least $g(x)=8-x$ . Use interval notation.
1. Hint 1
  Compare the rules by subtracting one from the other.
2. Hint 2
  The zeros of their difference divide the possible signs.
Answer

$(-\infty,\frac{-3-\sqrt{41}}2]\cup[\frac{-3+\sqrt{41}}2,\infty)$ .

Full solution

The comparison becomes
$x^2+3x-8\ge0.$
Its zeros are
$x=\frac{-3\pm\sqrt{41}}2$
The leading coefficient is positive, so the difference is nonnegative outside the two zeros.
Equality is allowed, so both endpoints are included.
At $x=0$ the difference is $-8$ , confirming exclusion of the middle interval.

Answer

$(-\infty,\frac{-3-\sqrt{41}}2]\cup[\frac{-3+\sqrt{41}}2,\infty)$ .

Key idea

Comparing two functions means finding the sign of their difference.
Problem 2: Mark complete
Problem 3 At the highest value

Solve $-3(x+1)^2+4\ge4$ over the reals.
1. Hint 1
  Consider the possible signs of a square and how a negative multiplier changes them.
2. Hint 2
  A nonnegative square can have a nonpositive value only at zero.
Answer

$\{-1\}$ .

Full solution

Subtract $4$ and divide by $-3$ , reversing the inequality.
$(x+1)^2\le0$
A square is nonnegative, so it must equal zero.
Thus $x=-1$ , which makes the original expression exactly $4$ .

Answer

$\{-1\}$ .

Key idea

At a quadratic’s maximum, an at-least-maximum condition can accept just its vertex input.
Problem 3: Mark complete
Problem 4 Two operating limits

A machine setting satisfies $0\le x\le5$ . Its output is $Q(x)=x(8-x)$ units, and operation requires $Q(x)\ge12$ . Find all allowed settings.
1. Hint 1
  Solve the output condition, then intersect with the setting domain.
2. Hint 2
  The quadratic boundary equation has two roots, but the domain may trim the resulting interval.
Answer

$[2,5]$ .

Full solution

The output condition rearranges to
$\begin{aligned} x^2-8x+12&\le0 \\ (x-2)(x-6)&\le0 \end{aligned}$

The upward quadratic $(x-2)(x-6)$ is negative between its roots and zero at them, so the non-strict inequality holds on $[2,6]$ .
Intersecting with the declared domain $[0,5]$ leaves $[2,5]$ .
At $x=2$ , output is $12$ ; at $x=5$ , it is $15$ , so both retained endpoints are allowed.

Answer

$[2,5]$ .

Key idea

A quadratic threshold must be intersected with a model’s permitted input domain.
Problem 4: Mark complete
Problem 5 Reading a level from a graph

The figure shows $y=f(x)$ and the horizontal line $y=4$ . Read the solution set of $f(x)<4$ in interval notation.

The graph of f, together with the horizontal line y = 4.

Text description of this figure

A grid with the horizontal axis running from negative 5 to 3 and the vertical axis running from negative 1 to 10, gridlines and number labels at every whole number. A single smooth upward opening curve, labeled f, enters through the top edge on the left, descends to its lowest point on the horizontal axis, and rises back up to leave through the top edge on the right. A dashed horizontal line, labeled y equals 4, crosses the full width of the grid, and the curve crosses this line at two points that are not marked.
1. Hint 1
  Compare the curve’s height with the horizontal reference line.
2. Hint 2
  The crossings with that line set the excluded endpoints.
Answer

$(-3,1)$ .

Full solution

The curve meets the reference line at horizontal coordinates $-3$ and $1$ .
Between them it lies below the line; outside them it lies above.
The strict inequality excludes the two crossing inputs, so the interval is $(-3,1)$ .

Answer

$(-3,1)$ .

Key idea

A function inequality with a nonzero threshold compares a graph with a horizontal line.
Problem 5: Mark complete
Problem 6 A vertical adjustment

For real $c$ , determine when $2x^2+4x+c<0$ has no real solution. Explain using the lowest value of the expression.
1. Hint 1
  Rewrite the varying terms as a positive multiple of a square.
2. Hint 2
  The strict inequality requires the minimum to lie below zero.
Answer

$c\ge2$ .

Full solution

Completing the square gives
$2(x+1)^2+c-2$
Its minimum is $c-2$ .
If $c\ge2$ , every value is at least zero and the strict inequality has no solution.
If $c<2$ , the input $x=-1$ makes the expression negative.
Thus the condition is exactly $c\ge2$ .

Answer

$c\ge2$ .

Key idea

A strict negative-value request is impossible exactly when an upward quadratic’s minimum is at least zero.
Problem 6: Mark complete
Problem 7 A difference of products

Solve $x(x+1)>2(x-1)^2$ over the reals, in interval notation.
1. Hint 1
  Move both expanded expressions to one side.
2. Hint 2
  Read the sign between the real zeros of the resulting quadratic.
Answer

$(\frac{5-\sqrt{17}}2,\frac{5+\sqrt{17}}2)$ .

Full solution

Expansion and rearrangement give
$\begin{aligned} x^2+x&>2x^2-4x+2 \\ x^2-5x+2&<0 \end{aligned}$

The roots are $(5-\sqrt{17})/2$ and $(5+\sqrt{17})/2$ .
The upward quadratic is negative between its roots; strictness excludes them.
The input $x=1$ works in the original, confirming the middle region.

Answer

$(\frac{5-\sqrt{17}}2,\frac{5+\sqrt{17}}2)$ .

Key idea

Collecting two products can reveal a quadratic whose sign decides their comparison.
Problem 7: Mark complete
Problem 8 One sample on an interval

A quadratic $f$ has no zero in $(1,4)$ , and $f(2)<0$ . A student says $f(3)$ must also be negative. Is that conclusion valid? Explain.
1. Hint 1
  A quadratic’s sign can change only at a zero.
2. Hint 2
  The two tested inputs lie in the same interval without a zero.
Answer

Yes; $f(3)<0$ .

Full solution

The inputs $2$ and $3$ lie inside $(1,4)$ .
A sign change between them would require a root somewhere in that interval, contrary to the hypothesis.
The sign therefore stays negative throughout the interval, and in particular at $3$ .

Answer

Yes; $f(3)<0$ .

Key idea

A quadratic keeps a fixed sign on an interval containing none of its zeros.
Problem 8: Mark complete
Problem 9 A pair of endpoints

A student solves $(x+3)(x-4)\ge0$ and writes $(-\infty,-3)\cup(4,\infty)$ . Evaluate the proposed answer and correct it if needed.
1. Hint 1
  Check whether the boundary inputs satisfy the original symbol.
2. Hint 2
  Keep the correctly chosen intervals and repair only excluded values that should be allowed.
Answer

Incorrect; $(-\infty,-3]\cup[4,\infty)$ .

Full solution

The product is positive outside its zeros and negative between them, so the outside pieces are correct.
At either root the product is $0$ .
Since equality is allowed, both roots must be included:
$\begin{aligned} x&\le-3 \\ \text{or}\quad x&\ge4 \end{aligned}$
so the corrected union has closed finite endpoints, $(-\infty,-3]\cup[4,\infty)$ .

Answer

Incorrect; $(-\infty,-3]\cup[4,\infty)$ .

Key idea

Non-strict inequalities include the zeros that form their finite interval endpoints.
Problem 9: Mark complete
Problem 10 One inequality implies another

A quadratic $f$ has positive leading coefficient and negative discriminant. Is $f(x)+2>0$ true for every real $x$ ? Justify.
1. Hint 1
  A negative discriminant rules out real zeros of the original quadratic.
2. Hint 2
  Use the opening direction to determine which sign the original function keeps.
Answer

Yes; $f(x)+2>0$ for every real $x$ .

Full solution

The original quadratic has no real zero and opens upward, so it is positive everywhere.
In vertex form its height is
$-\Delta/(4a)>0.$
Adding $2$ gives
$f(x)+2>2$
Thus the requested strict inequality holds for every real input.

Answer

Yes; $f(x)+2>0$ for every real $x$ .

Key idea

Adding a positive constant to an everywhere-positive quadratic preserves strict positivity.
Problem 10: Mark complete

← Back to lesson