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Quadratic Functions and Equations: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

0 of 10 completed

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Problem 1 of 10

Problem 1 One root survives a shift

Difficulty: 1 of 3 stars, Stretch

A monic real quadratic $q(x)$ has two distinct real roots. The equations $q(x)=0$ and $q(x+3)=0$ have exactly one common real solution, and $q(1)=10$ . Find every possible polynomial $q$ , and identify the common solution in each case.
1. Hint 1
  Compare the two root sets. How does replacing $x$ by $x+3$ move each root?
2. Hint 2
  Write the smaller root as $r$ . The shared-root condition forces the larger root to be $r+3$ ; now use $q(1)$ .
Answer

$q(x)=x^2-9x+18$ , with common solution $3$ , or $q(x)=x^2+5x+4$ , with common solution $-4$ .

Full solution

Let the roots be $r<s$ .
The roots of $q(x+3)$ are $r-3$ and $s-3$ .
The number $r-3$ lies below both roots of $q$ , so a common root is possible only when $s-3=r$ .
Thus $s=r+3$ .
Conversely, this separation gives exactly one shared root, namely $r$ .

Monicity now gives
$q(x)=(x-r)(x-r-3).$
Substituting $x=1$ yields $(1-r)(-2-r)=10$ , or
$r^2+r-12=0.$
Therefore $(r+4)(r-3)=0$ , so $r=3$ or $r=-4$ .
The resulting polynomials are $(x-3)(x-6)$ and $(x+4)(x+1)$ .
Both have distinct roots three units apart, both give $10$ at $x=1$ , and the preceding root-set argument verifies exactly one common solution for each.

Answer

$q(x)=x^2-9x+18$ , with common solution $3$ , or $q(x)=x^2+5x+4$ , with common solution $-4$ .

Key idea

Translate a shifted equation into a shifted root set before comparing coefficients.
Problem 1: Mark complete
Problem 2 How far can the next value move?

Difficulty: 1 of 3 stars, Stretch

A real quadratic $q$ satisfies $0\le q(0),q(1),q(2)\le1$ . Determine the least and greatest possible values of $q(3)$ . Prove both bounds and identify every quadratic attaining either bound.

Builds on Linear Inequalities
1. Hint 1
  Express $q(3)$ using only the three known values. The coefficients of $x^2$ , $x$ , and $1$ must each match.
2. Hint 2
  Check the identity $q(3)=q(0)-3q(1)+3q(2)$ , and consider which choices make each term largest or smallest.
Answer

The range is $[-3,4]$ . The minimum occurs only for $q(x)=2x-x^2$ ; the maximum occurs only for $q(x)=(x-1)^2$ .

Full solution

Writing $q(x)=ax^2+bx+c$ and collecting coefficients verifies
$q(3)=q(0)-3q(1)+3q(2).$
Each of the three values belongs to $[0,1]$ , so
$-3\le q(3)\le4.$
Equality on the right requires $q(0)=1$ , $q(1)=0$ , and $q(2)=1$ : any failure makes at least one of the three bounded terms smaller.
These values determine
$q(x)=(x-1)^2.$
Equality on the left requires the values $0,1,0$ , giving
$q(x)=2x-x^2.$

Three values determine a quadratic uniquely because subtracting two candidates gives a polynomial of degree at most two with three distinct zeros; alternatively, solving the three coefficient equations gives the displayed quadratics directly.
To verify that every intermediate value is possible, use $q_t(x)=(1-t)(2x-x^2)+t(x-1)^2$ , for
$0\le t\le1.$
It gives
$q_t(3)=-3+7t.$
At $t=1/2$ this expression becomes the constant $1/2$ , but the same output $1/2$ is obtained by the genuine quadratic
$q(x)=(x-1)^2/8.$
Thus the whole reported interval is attainable by quadratics.

Answer

The range is $[-3,4]$ . The minimum occurs only for $q(x)=2x-x^2$ ; the maximum occurs only for $q(x)=(x-1)^2$ .

Key idea

A signed linear combination of known values can give a sharp bound at an unmeasured input.
Problem 2: Mark complete
Problem 3 Between integer inputs

Difficulty: 1 of 3 stars, Stretch

A monic quadratic $q(x)$ has real coefficients and satisfies $q(n)\ge0$ for every integer $n$ . Find the smallest value that $q(x)$ can possibly take at a real input, over all such quadratics. Identify every polynomial attaining this sharp lower bound somewhere, and prove it satisfies the condition at every integer.

Builds on Completing the Square
1. Hint 1
  Use vertex form $q(x)=(x-h)^2+m$ . There is always an integer within $1/2$ of $h$ .
2. Hint 2
  If $q$ reaches the proposed sharp bound, the nearest integer must be exactly $1/2$ from the vertex. What does that say about the two adjacent integers?
Answer

The sharp bound is $-1/4$ . Exactly the polynomials $q(x)=(x-k)(x-k-1)$ , with $k$ an integer, attain it.

Full solution

Write $q(x)=(x-h)^2+m$ , so its minimum over real inputs is $m$ .
Choose a nearest integer $n$ to $h$ ; one exists with
$|n-h|\le1/2.$
Since $q(n)\ge0$ , we have
$m\ge-(n-h)^2\ge-1/4.$
This proves a universal bound, even though the hypothesis checks only integer inputs.

If the minimum is $-1/4$ , the nearest-integer distance must equal $1/2$ .
Therefore $h=k+1/2$ for some integer $k$ , and
$q(x)=(x-k-1/2)^2-1/4=(x-k)(x-k-1).$
Conversely, at any integer $n$ the two factors $n-k$ and $n-k-1$ are consecutive integers.
They are either both nonnegative, both nonpositive, or one is zero, so their product is nonnegative.
Each such polynomial reaches $-1/4$ at $x=k+1/2$ .
Thus both the sharp value and all equality cases have been established.

Answer

The sharp bound is $-1/4$ . Exactly the polynomials $q(x)=(x-k)(x-k-1)$ , with $k$ an integer, attain it.

Key idea

A nearby discrete test point can control a continuous minimum, with the distance to the test grid setting the error.
Problem 3: Mark complete
Problem 4 When square outputs keep appearing

Difficulty: 2 of 3 stars, Challenge

(a) Find every nonnegative integer $n$ for which $n^2+11n+24$ is a perfect square.

(b) Let $b,c$ be fixed integers. Prove that $n^2+bn+c$ is a perfect square for infinitely many nonnegative integers $n$ if and only if $b$ is even and $c=b^2/4$ . Here a perfect square means the square of an integer, including $0$ .

Builds on Completing the Square
1. Hint 1
  Complete a square after multiplying the equation by $4$ , so that every quantity remains an integer.
2. Hint 2
  If $n^2+bn+c=m^2$ , then $(2n+b-2m)(2n+b+2m)=b^2-4c$ . A fixed nonzero integer has only finitely many integer factor pairs.
Answer

(a) Only $n=1$ , giving $36$ . (b) Exactly the quadratics $(n+k)^2$ , with integer $k$ , have infinitely many such outputs.

Full solution

For part (a), choose $m\ge0$ with
$n^2+11n+24=m^2.$
Completing the square gives
$(2n+11-2m)(2n+11+2m)=25.$
Their sum is $4n+22>0$ , so the two factors are positive; the first is no greater than the second.
The only ordered possibilities are $(1,25)$ and $(5,5)$ .
Their sums give $n=1$ and $n=-3$ , respectively.
The nonnegative restriction leaves only $n=1$ , and direct substitution gives $36$ .

For part (b), any square output satisfies $(2n+b-2m)(2n+b+2m)=D$ , where $D=b^2-4c$ is fixed.
If $D\ne0$ , each first factor is a nonzero integer divisor of $D$ and has absolute value at most $|D|$ .
There are only finitely many factor pairs, and each determines at most one $n$ from their sum $4n+2b$ .
Thus infinitely many square outputs force $D=0$ .
Then $b^2=4c$ implies $b$ is even, and the polynomial is $(n+b/2)^2$ .
Conversely, this form is the square of an integer for every nonnegative integer $n$ .
This proves both directions.

Answer

(a) Only $n=1$ , giving $36$ . (b) Exactly the quadratics $(n+k)^2$ , with integer $k$ , have infinitely many such outputs.

Key idea

Completing the square can turn a claim about infinitely many integer inputs into a fixed-product obstruction.
Problem 4: Mark complete
Problem 5 Can the roots make a triangle?

Difficulty: 2 of 3 stars, Challenge

The real roots $a\le b$ of $x^2-10x+k=0$ are to be two side lengths of a nondegenerate triangle whose third side is $6$ .

(a) Find exactly which real values of $k$ permit such a triangle.

(b) Determine the area in terms of $k$ , and find its largest possible value and all side lengths attaining that value. Use an algebraic or coordinate argument; a triangle-area formula in terms of three sides may not be assumed.

Schematic; side lengths vary with $k$ .

Text description of this figure

A schematic triangle with a dot at each vertex. Its horizontal base is labeled 6. The top vertex sits to the left of center, so the left side, labeled a, is drawn shorter than the right side, labeled b. The drawing is not to scale: the side lengths a and b vary with k.
1. Hint 1
  The sum $a+b$ is fixed. Translate the triangle inequalities into a condition on $b-a$ .
2. Hint 2
  Place the side of length $6$ from $(0,0)$ to $(6,0)$ , and the third vertex at $(t,h)$ . Subtract the two distance-squared equations to find $t$ .
Answer

(a) $16<k\le25$ . (b) The area is $4\sqrt{k-16}$ ; its maximum is $12$ , attained exactly by side lengths $5,5,6$ .

Full solution

The root sum is $a+b=10$ , and real roots require $k\le25$ .
Put
$d=b-a=\sqrt{100-4k}\ge0.$
A triangle with sides $a,b,6$ exists exactly when $a+b>6$ and $b-a<6$ , together with positivity.
The first inequality is automatic, and $d<6$ implies $a=(10-d)/2>2$ , so positivity also follows.
Therefore the exact condition is $0\le d<6$ , equivalently $16<k\le25$ .
The strict lower bound excludes a flattened triangle.

For the area calculation, place the third vertex at $(t,h)$ with $h>0$ , its distances from $(0,0)$ and $(6,0)$ being $a$ and $b$ .
Subtracting $t^2+h^2=a^2$ and $(t-6)^2+h^2=b^2$ gives $t=3-5d/6$ .
Hence
$h^2=a^2-t^2=16-4d^2/9=16(k-16)/9.$
The area is
$6h/2=4\sqrt{k-16}.$
This expression increases as $k$ increases over the feasible interval, so its maximum is $12$ at $k=25$ , when both roots equal $5$ .
The positive height formula also constructs a triangle for every permitted $k$ .

Answer

(a) $16<k\le25$ . (b) The area is $4\sqrt{k-16}$ ; its maximum is $12$ , attained exactly by side lengths $5,5,6$ .

Key idea

Root sums and differences translate algebraic conditions into geometric feasibility before optimization.
Problem 5: Mark complete
Problem 6 The smaller of two scores

Difficulty: 2 of 3 stars, Challenge

You may choose a real number $x$ with $2\le x\le12$ . Your guaranteed score is the smaller of $A=x(12-x)$ and $B=(x-2)(14-x)$ .

(a) Find all choices guaranteeing a score of at least $32$ .

(b) Find the largest possible guaranteed score, and prove which choices attain it.

Builds on Quadratic Inequalities
1. Hint 1
  A guarantee of at least $32$ requires both scores to be at least $32$ . For the maximum, first determine where the two scores exchange order.
2. Hint 2
  Compute $A-B=28-4x$ . On the side where $B$ is smaller, compare $B$ with $35$ ; do the analogous comparison for $A$ on the other side.
Answer

(a) $6\le x\le8$ . (b) The largest guaranteed score is $35$ , attained only at $x=7$ .

Full solution

For part (a), $A\ge32$ is equivalent to $(x-4)(x-8)\le0$ , so
$4\le x\le8.$
Similarly, $B\ge32$ becomes $(x-6)(x-10)\le0$ , so
$6\le x\le10.$
The two conditions hold together exactly on $[6,8]$ , which lies inside the permitted domain.

For part (b), $A-B=28-4x$ .
Thus the smaller score is $B$ when $x\le7$ and $A$ when $x\ge7$ .
If $2\le x\le7$ , then $B-35=-x^2+16x-63=-(x-7)(x-9)\le0$ , with equality in this interval only at $7$ .
If $7\le x\le12$ , then $A-35=-x^2+12x-35=-(x-5)(x-7)\le0$ , again with equality only at $7$ .
At $x=7$ , both scores are $35$ .
These bounds cover the full domain, so the guarantee cannot be improved by choosing a value away from the crossing point.

Answer

(a) $6\le x\le8$ . (b) The largest guaranteed score is $35$ , attained only at $x=7$ .

Key idea

Maximizing a guaranteed outcome often means splitting the domain according to which competing quantity is smaller.
Problem 6: Mark complete
Problem 7 Coefficients for a prescribed root window

Difficulty: 2 of 3 stars, Challenge

For each real number $s$ , determine every real $p$ for which both roots of $x^2-sx+p=0$ , counted with multiplicity, belong to the interval $[1,4]$ . Your answer must include the exact allowed range of $s$ , sharp lower and upper bounds for $p$ , and the roots at each bound.

Builds on Sum and Product of Roots, Completing the Square
1. Hint 1
  Write the roots as $a$ and $s-a$ , with $a\le s-a$ . Which interval of $a$ keeps both roots in $[1,4]$ ?
2. Hint 2
  The product is $p=s^2/4-(a-s/2)^2$ . The smallest allowable root is $\max(1,s-4)$ .
Answer

For $2\le s\le5$ , $s-1\le p\le s^2/4$ ; for $5\le s\le8$ , $4s-16\le p\le s^2/4$ . No other $s$ works. The upper bound has roots $s/2,s/2$ ; the lower bound has roots $1,s-1$ or $s-4,4$ , respectively.

Full solution

Let $a\le b$ be the roots.
If $1\le a,b\le4$ , then
$2\le s=a+b\le8.$
For such an $s$ , the conditions on the smaller root are exactly $\max(1,s-4)\le a\le s/2$ , with $b=s-a$ .
The interval is nonempty throughout $[2,8]$ .
The product satisfies $p=a(s-a)=s^2/4-(a-s/2)^2$ , so it is largest at $a=s/2$ .
It is smallest at the left endpoint of the allowed interval.

When $2\le s\le5$ , that left endpoint is $1$ , giving $p\ge s-1$ .
When $5\le s\le8$ , it is $s-4$ , giving
$p\ge4(s-4).$
These values give the stated endpoint roots.
For completeness, any $p$ between the lower bound $L$ and $s^2/4$ yields $a=s/2-\sqrt{s^2/4-p}$ between $\max(1,s-4)$ and $s/2$ , and $b=s-a$ in $[1,4]$ .
Their sum and product are $s,p$ , so they are exactly the roots of the required quadratic.
This proves sufficiency as well as necessity, including repeated-root and endpoint cases.

Answer

For $2\le s\le5$ , $s-1\le p\le s^2/4$ ; for $5\le s\le8$ , $4s-16\le p\le s^2/4$ . No other $s$ works. The upper bound has roots $s/2,s/2$ ; the lower bound has roots $1,s-1$ or $s-4,4$ , respectively.

Key idea

For a fixed sum, the product is controlled by how far the two roots can separate inside their permitted interval.
Problem 7: Mark complete
Problem 8 A line that tracks a parabola

Difficulty: 3 of 3 stars, Deep challenge

Choose a line $L(x)=ax+b$ to approximate $x^2$ throughout the interval $0\le x\le4$ . Let $E$ be the smallest number for which $|x^2-L(x)|\le E$ at every input in that interval. Find the least possible $E$ over all lines, prove the optimum holds over the entire interval, and prove the optimizing line is unique.
1. Hint 1
  Look at the errors at $x=0,2,4$ . A particular weighted combination cancels the line completely.
2. Hint 2
  If $e(x)=x^2-ax-b$ , then $e(0)-2e(2)+e(4)=8$ . How large must one of the three absolute errors be?
Answer

The least possible error is $E=2$ , and the unique optimizing line is $L(x)=4x-2$ .

Full solution

Let
$e(x)=x^2-ax-b.$
The identity $e(0)-2e(2)+e(4)=8$ holds for every choice of $a,b$ .
If $|e(x)|\le E$ throughout the interval, then $8\le |e(0)|+2|e(2)|+|e(4)|\le4E$ , so necessarily $E\ge2$ .
This uses only three inputs, but gives a lower bound for every possible line.

The line $L(x)=4x-2$ has error
$e(x)=x^2-4x+2=(x-2)^2-2.$
Because $0\le(x-2)^2\le4$ on $[0,4]$ , its error belongs to $[-2,2]$ at every input.
Thus $E=2$ is achievable.
Finally, equality in the three-input bound requires $e(0)=2$ , $e(2)=-2$ , and $e(4)=2$ : each signed term in $e(0)-2e(2)+e(4)$ must reach its individual maximum.
The first condition gives $b=-2$ , and the middle one gives $a=4$ .
Therefore no other line achieves the same best error.

Answer

The least possible error is $E=2$ , and the unique optimizing line is $L(x)=4x-2$ .

Key idea

A few carefully weighted test points can certify a global approximation bound, while alternating extreme errors can force uniqueness.
Problem 8: Mark complete
Problem 9 Three numbers with two measurements

Difficulty: 3 of 3 stars, Deep challenge

Real numbers $x,y,z$ satisfy $x+y+z=9$ and $x^2+y^2+z^2=33$ . Find the least and greatest possible values of $xyz$ , and identify every triple attaining either extreme. Prove your bounds without calculus or a general formula for cubic roots.
1. Hint 1
  Center the numbers at their common average: write $a=x-3$ , $b=y-3$ , and $c=z-3$ .
2. Hint 2
  From $a+b=-c$ and $a^2+b^2+c^2=6$ , obtain $ab=c^2-3$ and $|c|\le2$ . Factor $c^3-3c-2$ and $c^3-3c+2$ .
Answer

The minimum is $16$ , attained by permutations of $(1,4,4)$ . The maximum is $20$ , attained by permutations of $(2,2,5)$ .

Full solution

Set $a=x-3$ , $b=y-3$ , and $c=z-3$ .
The constraints become $a+b+c=0$ and
$a^2+b^2+c^2=6.$
Squaring the zero sum gives $ab+bc+ca=-3$ , so
$xyz=(a+3)(b+3)(c+3)=18+abc.$
Also, $(a+b)^2=c^2$ and $a^2+b^2=6-c^2$ give $ab=c^2-3$ .
Therefore
$abc=c^3-3c.$

The two numbers $a,b$ are real only if $(a-b)^2=12-3c^2\ge0$ , hence
$-2\le c\le2.$
On this interval, $c^3-3c-2=(c-2)(c+1)^2\le0$ and
$c^3-3c+2=(c+2)(c-1)^2\ge0.$
Thus $-2\le abc\le2$ , proving
$16\le xyz\le20.$
Equality $abc=2$ forces $c=2$ or $c=-1$ ; using the sum and product of $a,b$ then gives exactly permutations of
$(a,b,c)=(2,-1,-1).$
Equality $abc=-2$ similarly gives permutations of $(-2,1,1)$ .
Shifting back yields the claimed triples.
Direct substitution verifies their two measurements and products.

Answer

The minimum is $16$ , attained by permutations of $(1,4,4)$ . The maximum is $20$ , attained by permutations of $(2,2,5)$ .

Key idea

Centering symmetric constraints can reduce a three-variable extremum to a one-variable factorization with a tightly controlled domain.
Problem 9: Mark complete
Problem 10 An integer equation with no last solution

Difficulty: 3 of 3 stars, Deep challenge

Find all ordered pairs of positive integers $(x,y)$ satisfying $x^2+y^2+5=5xy$ . A complete answer may use a precisely defined recursive construction, but you must prove that it produces only solutions and that every solution appears. Give the first four pairs with $x\le y$ in each of your resulting families.
1. Hint 1
  For a fixed $x$ , regard the equation as a quadratic in $y$ . What is its other root?
2. Hint 2
  For $2\le x\le y$ , show that the other root $5x-y$ is a positive integer smaller than $x$ . This gives a descent to small first coordinates.
Answer

Up to swapping coordinates, there are two families: start at $(1,2)$ or $(1,3)$ and repeatedly replace $(u,v)$ by $(v,5v-u)$ . Their first four pairs are $(1,2),(2,9),(9,43),(43,206)$ and $(1,3),(3,14),(14,67),(67,321)$ .

Full solution

Order a solution so that $x\le y$ .
With $x$ fixed, $y$ is a root of
$Y^2-5xY+x^2+5=0.$
Its other root is the integer $y_*=5x-y$ , and the product identity $yy_*=x^2+5$ shows $y_*>0$ .
For $x\ge2$ , the quadratic evaluated at $Y=x$ is $5-3x^2<0$ .
An upward-opening quadratic is negative strictly between its real roots, so $y_*<x<y$ .
Replacing the ordered pair by $(y_*,x)$ gives a new positive integer solution with a strictly smaller largest coordinate.

This descent must terminate, and it can terminate only when the smaller coordinate is $1$ .
The equation then gives $y^2-5y+6=0$ , so the terminal pair is $(1,2)$ or $(1,3)$ .
Reversing a descent step sends $(u,v)$ to $(v,5v-u)$ .
Substitution, or the other-root relation, proves this step preserves the equation.
Starting from either base pair, $v>u>0$ implies $5v-u>v$ , so every later pair remains positive and ordered.
Every original ordered solution is obtained by reversing its finite descent; swaps give every ordered pair.
Applying the recurrence produces the four displayed pairs in each family.
The unique decreasing step above $x=1$ also shows the two families cannot merge.

Answer

Up to swapping coordinates, there are two families: start at $(1,2)$ or $(1,3)$ and repeatedly replace $(u,v)$ by $(v,5v-u)$ . Their first four pairs are $(1,2),(2,9),(9,43),(43,206)$ and $(1,3),(3,14),(14,67),(67,321)$ .

Key idea

An integer quadratic can carry its own descent: replacing one root by the other may shrink a solution while preserving the equation.
Problem 10: Mark complete