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Quadratic 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.

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Problem 1 of 10

Problem 1 A new equation from old roots

Difficulty: 1 of 3 stars, Stretch

The real numbers $r$ and $s$ are the roots of $2x^2-7x+3=0$ . Define $u=1/(r-1)$ and $v=1/(s-1)$ .

Using sums and products of roots, construct a quadratic equation with integer coefficients whose roots are $u$ and $v$ . Explain why the definitions are valid and why exactly one of $u,v$ is negative.

Builds on Sums and Products of Roots, Algebraic Fractions
1. Hint 1
  What are $r+s$ and $rs$ ? Express $(r-1)(s-1)$ using those two quantities.
2. Hint 2
  Find $u+v$ and $uv$ , then use $(y-u)(y-v)=y^2-(u+v)y+uv$ .
Answer

$2y^2+3y-2=0$ ; exactly one transformed root is negative.

Full solution

The original equation gives $r+s=7/2$ and $rs=3/2$ .
Also, $x=1$ is not a root: substitution gives $2-7+3=-2$ .
Thus neither denominator in the definitions of $u,v$ is zero.

Combine the denominators before trying to find either root:
$(r-1)(s-1)=rs-(r+s)+1=-1.$
Consequently, $uv=-1$ and
$u+v=\frac{r+s-2}{(r-1)(s-1)}=-\frac32.$

The monic equation for the new roots is
$y^2+\frac32y-1=0.$
Multiplication by $2$ gives the requested integer coefficients.
Since $u,v$ are real and have negative product, they are both nonzero and have opposite signs.
As a check, the equation factors as $(2y-1)(y+2)=0$ , giving $1/2$ and $-2$ .

Answer

$2y^2+3y-2=0$ ; exactly one transformed root is negative.

Key idea

Transform the sum and product together; individual roots are often unnecessary.
Problem 1: Mark complete
Problem 2 When the constant follows the sum

Difficulty: 1 of 3 stars, Stretch

Find all positive integers $m$ for which $x^2-mx+3m=0$ has two positive integer roots. The two roots are allowed to be equal. Prove that your list is complete.

Builds on Sums and Products of Roots, Factoring Quadratics
1. Hint 1
  Call the roots $r,s$ . What equation follows from $r+s=m$ and $rs=3m$ ?
2. Hint 2
  Rewrite $rs=3r+3s$ as a product involving $r-3$ and $s-3$ . First determine the signs of these factors.
Answer

$m=12$ with roots $6,6$ , or $m=16$ with roots $4,12$ .

Full solution

If the positive integer roots are $r,s$ , their sum is $m$ and their product is $3m$ .
Therefore $rs=3r+3s$ , or
$(r-3)(s-3)=9.$
The signs of these new factors matter.
Dividing $rs=3r+3s$ by $3rs$ gives
$1/r+1/s=1/3.$
Since both summands are positive, each is less than $1/3$ , so $r>3$ and $s>3$ .

Thus $r-3$ and $s-3$ are positive integers with product $9$ .
Up to order, the only factor pairs are $(1,9)$ and $(3,3)$ .
They give $(r,s)=(4,12)$ and $(6,6)$ , respectively.
Hence $m=r+s$ is $16$ or $12$ .

Both candidates work: $x^2-16x+48=(x-4)(x-12)$ and
$x^2-12x+36=(x-6)^2.$
The positive factor-pair list proves that there are no other possibilities; the repeated-root case must be retained.

Answer

$m=12$ with roots $6,6$ , or $m=16$ with roots $4,12$ .

Key idea

A relation between a sum and a product can become a finite factor-pair problem after a shift.
Problem 2: Mark complete
Problem 3 Exactly one surviving root

Difficulty: 1 of 3 stars, Stretch

For a real parameter $k$ , consider
$\frac{x^2-5x+6}{x-2}=\frac{k}{x-3}.$
Find every value of $k$ for which the equation has exactly one distinct real solution, and identify that solution. Justify your answer by classifying the number of solutions for every real $k$ .

Builds on Factoring Quadratics, Algebraic Fractions
1. Hint 1
  Write down the forbidden values of $x$ before canceling a factor. What simpler equation is equivalent on the allowed domain?
2. Hint 2
  The reduced equation is $(x-3)^2=k$ , with $x\ne2,3$ . Separate negative, zero, and positive $k$ , then ask when a candidate is forbidden.
Answer

Only $k=1$ , with sole solution $x=4$ . There are no solutions for $k\le0$ and two for $k>0$ , $k\ne1$ .

Full solution

The original fractions require $x\ne2,3$ , even when $k=0$ .
Factoring the numerator as $(x-2)(x-3)$ and canceling on this domain gives
$x-3=k/(x-3).$
Multiplication by the nonzero denominator yields $(x-3)^2=k$ .
Every step is reversible for allowed $x$ .

If $k<0$ , a real square cannot equal $k$ , so there are no solutions.
If $k=0$ , the reduced equation has only the candidate $x=3$ , which is forbidden.
Again there are no solutions.

If $k>0$ , the reduced equation has two distinct candidates $3+\sqrt{k}$ and $3-\sqrt{k}$ .
Neither equals $3$ .
The larger candidate exceeds $3$ and therefore cannot equal $2$ .
The smaller candidate equals $2$ precisely when $\sqrt{k}=1$ , that is, when $k=1$ .
Thus this is the only positive parameter that removes one candidate.
It leaves $x=4$ , which satisfies the original equation.
Every other positive $k$ gives two allowed roots.
This covers all real parameters.

Answer

Only $k=1$ , with sole solution $x=4$ . There are no solutions for $k\le0$ and two for $k>0$ , $k\ne1$ .

Key idea

A parameter can change the number of solutions by merging roots or by moving a root to an excluded input.
Problem 3: Mark complete
Problem 4 How many different roots?

Difficulty: 2 of 3 stars, Challenge

For a real parameter $t$ , consider the two equations
$x^2-(t+4)x+4t=0,\qquad x^2-(2t+1)x+2t=0.$
Find every value of $t$ for which exactly three distinct real numbers solve at least one of these equations. Prove that for all other values there are exactly four such numbers.

Builds on Factoring Quadratics, Sums and Products of Roots
1. Hint 1
  Factor each quadratic using its parameter. Count distinct numbers, not roots with multiplicity.
2. Hint 2
  The four entries to compare are $4,t,1,2t$ . List every possible equality between two entries and then check each exceptional parameter.
Answer

$t\in\{0,\tfrac12,1,2,4\}$ gives exactly three distinct numbers; every other real $t$ gives four.

Full solution

The equations factor as $(x-4)(x-t)=0$ and
$(x-1)(x-2t)=0.$
Their combined list of roots is therefore $4,t,1,2t$ .
A reduction from four distinct numbers occurs precisely when two entries agree.

There are six pairs to consider.
The equality $4=1$ is impossible.
The other five equalities give $t=4$ from $4=t$ ; $t=2$ from $4=2t$ ; $t=1$ from $t=1$ ; $t=0$ from $t=2t$ ; and $t=1/2$ from $1=2t$ .
This checks coincidences both within one quadratic and between the two quadratics.

The five exceptional parameters give the distinct-root sets $\{0,1,4\}$ , $\{1/2,1,4\}$ , $\{1,2,4\}$ , $\{1,2,4\}$ , and $\{1,4,8\}$ for $t=0,1/2,1,2,4$ , respectively.
Each has exactly three elements.
Outside this list, none of the six pairs agrees, so all four entries are distinct.
In particular, there is no parameter producing only one or two distinct roots.

Answer

$t\in\{0,\tfrac12,1,2,4\}$ gives exactly three distinct numbers; every other real $t$ gives four.

Key idea

Count repeated roots inside each equation as well as shared roots between equations.
Problem 4: Mark complete
Problem 5 A fraction without the roots

Difficulty: 2 of 3 stars, Challenge

The distinct real roots of $3x^2-12x+5=0$ are $a$ and $b$ . Without calculating either root, evaluate
$\frac{a}{b+1}+\frac{b}{a+1}.$
Justify that both denominators are nonzero.

Builds on Sums and Products of Roots, Algebraic Fractions
1. Hint 1
  Combine the two fractions. Which symmetric expressions occur in the numerator and denominator?
2. Hint 2
  Use $a^2+b^2=(a+b)^2-2ab$ , together with the root sum and product.
Answer

$5/2$ .

Full solution

The root relationships give $a+b=4$ and $ab=5/3$ .
Neither root is $-1$ , because substitution into the original expression gives
$3+12+5=20\ne0.$
Thus both fractions are defined.

Use a common denominator:
$\frac{a}{b+1}+\frac{b}{a+1}=\frac{a(a+1)+b(b+1)}{(a+1)(b+1)}.$
The numerator is $a^2+b^2+a+b$ .
Since $a^2+b^2=(a+b)^2-2ab=16-10/3=38/3$ , the numerator is $50/3$ .
The denominator is
$ab+a+b+1=5/3+4+1=20/3.$
Their ratio is $5/2$ .

The answer is unchanged if the roots are swapped.
That symmetry explains why their sum and product suffice: combining the fractions removes the need to distinguish the roots individually.
Keeping the denominator in symmetric form also provides a second check that it is nonzero.

Answer

$5/2$ .

Key idea

A symmetric rational expression may simplify to rational data even when its roots are irrational.
Problem 5: Mark complete
Problem 6 Coefficients that are also roots

Difficulty: 2 of 3 stars, Challenge

Find all ordered pairs of real numbers $(a,b)$ for which the roots of $x^2+ax+b=0$ , counted with multiplicity, are exactly $a$ and $b$ . Include all cases where a coefficient is zero, and prove completeness.

Builds on Sums and Products of Roots, Factoring Quadratics
1. Hint 1
  If $a,b$ really are the roots, their sum and product must match the coefficients.
2. Hint 2
  The product condition gives $b(a-1)=0$ . Treat the two factors separately before dividing by anything.
Answer

$(a,b)=(0,0)$ or $(1,-2)$ .

Full solution

The proposed roots have sum $a+b$ and product $ab$ .
Comparing with the equation gives $a+b=-a$ and $ab=b$ .
The first condition says $b=-2a$ ; the second factors as $b(a-1)=0$ .

If $b=0$ , then $b=-2a$ forces $a=0$ .
If $a=1$ , then $b=-2$ .
These cases exhaust the possibilities because a product of two real numbers is zero only when at least one factor is zero.
Dividing by $b$ at the outset would incorrectly lose the first case.

Finally, test sufficiency.
For $(0,0)$ , the equation is $x^2=0$ , whose two roots counted with multiplicity are $0,0$ .
For $(1,-2)$ , it is $x^2+x-2=(x-1)(x+2)=0$ , with roots $1,-2$ .
Both ordered pairs satisfy the requested coefficient-root correspondence.
The reversed pair $(-2,1)$ is not another solution: reversing coefficients changes the equation.

Answer

$(a,b)=(0,0)$ or $(1,-2)$ .

Key idea

When unknown coefficients are constrained by their own roots, use both root relationships and preserve zero cases.
Problem 6: Mark complete
Problem 7 Reconstruct the two equations

Difficulty: 2 of 3 stars, Challenge

The equations $x^2-ax+12=0$ and $x^2-bx+18=0$ have a common real root, and $a+b=17$ . Find all ordered pairs $(a,b)$ and the common root for each. Prove that your reconstructions work and that none is missing.

Builds on Sums and Products of Roots, Factoring Harder Quadratics
1. Hint 1
  Call the common root $r$ . Its value cannot be zero. Express each other root using $r$ .
2. Hint 2
  The coefficient condition becomes $2r+30/r=17$ . Factor the resulting quadratic in $r$ , then reconstruct both equations.
Answer

$(a,b)=(8,9)$ with common root $6$ , or $(a,b)=(73/10,97/10)$ with common root $5/2$ .

Full solution

Let $r$ be a common root.
It is nonzero because both constant terms are nonzero.
The other roots must be $12/r$ and $18/r$ .
Taking root sums yields $a=r+12/r$ and $b=r+18/r$ .
Hence $a+b=17$ gives $2r+30/r=17$ , or
$2r^2-17r+30=(2r-5)(r-6)=0.$

For $r=6$ , the other roots are $2$ and $3$ , giving $a=8,b=9$ .
For $r=5/2$ , the other roots are $24/5$ and $36/5$ , giving
$a=73/10,b=97/10.$
These are the only candidates produced by the factored equation.

Conversely, the pairs of factorizations $(x-6)(x-2)$ and $(x-6)(x-3)$ verify the first reconstruction.
The products $(x-5/2)(x-24/5)$ and $(x-5/2)(x-36/5)$ verify the second.
Their constant terms and coefficient sums are correct.
In each reconstruction the other roots differ from one another and from the common root, so the stated common root is unique.

Answer

$(a,b)=(8,9)$ with common root $6$ , or $(a,b)=(73/10,97/10)$ with common root $5/2$ .

Key idea

Represent the unknown equations by a common root and use products to recover their missing roots.
Problem 7: Mark complete
Problem 8 Which expression is larger?

Difficulty: 3 of 3 stars, Deep challenge

Let $r<s$ be the two real roots of $x^2-9x+17=0$ . Do not calculate the roots or use decimal approximations.

(a) Locate each root between consecutive integers.

(b) Determine whether $r^2$ or $s+1$ is larger. Give an exact argument.

Builds on Sums and Products of Roots, Inequality Basics
1. Hint 1
  For any real $c$ , $(r-c)(s-c)=c^2-9c+17$ . Its sign tells you whether $c$ lies between the roots.
2. Hint 2
  Use $r^2=9r-17$ and $s=9-r$ to reduce part (b) to comparing $r$ with $27/10$ . Evaluate the factored expression at that number.
Answer

(a) $2<r<3$ and $6<s<7$ . (b) $r^2<s+1$ .

Full solution

The root sum is $9$ and the product is $17$ .
At $c=3$ , $(r-3)(s-3)=9-27+17=-1$ , so $r<3<s$ .
At $c=2$ ,
$(r-2)(s-2)=4-18+17=3>0.$
Since $s>3>2$ , its second factor is positive; its first factor must therefore be positive as well.
Thus $2<r<3$ .
Using $s=9-r$ gives $6<s<7$ .

For the comparison, substitute the original equation for $r^2$ :
$r^2-(s+1)=(9r-17)-(10-r)=10r-27.$
We must decide which side of $27/10$ contains $r$ .

At $c=27/10$ ,
$(r-c)(s-c)=c^2-9c+17=-\frac{1}{100}<0.$
A negative product places $c$ strictly between $r$ and $s$ .
Hence $r<27/10$ , so $10r-27<0$ and $r^2<s+1$ .
The comparison is close, but the exact sign of a small rational number settles it without approximating either root.

Answer

(a) $2<r<3$ and $6<s<7$ . (b) $r^2<s+1$ .

Key idea

Evaluate a factored quadratic at a comparison point to locate roots without solving for them.
Problem 8: Mark complete
Problem 9 A reciprocal constraint

Difficulty: 3 of 3 stars, Deep challenge

Positive real numbers $a,b$ satisfy $1/a+1/b=1/2$ .

(a) Prove that $a^2+b^2\ge32$ , and determine exactly when equality holds.

(b) Find all ordered pairs $(a,b)$ satisfying the additional condition $a^2+b^2=45$ .

Builds on Sums and Products of Roots, Factoring Quadratics, Inequality Basics
1. Hint 1
  Write $s=a+b$ and $p=ab$ . The reciprocal condition links $s$ and $p$ .
2. Hint 2
  Use $(a-b)^2=s^2-4p\ge0$ to bound $s$ . For part (b), use $a^2+b^2=s^2-2p$ and factor.
Answer

(a) Equality only at $(a,b)=(4,4)$ . (b) $(3,6)$ and $(6,3)$ .

Full solution

Put $s=a+b$ and $p=ab$ .
Positivity gives $s,p>0$ , and the reciprocal equation gives $s/p=1/2$ , so $p=2s$ .
The nonnegative square $(a-b)^2=s^2-4p$ therefore yields
$s(s-8)\ge0.$
Since $s>0$ , we obtain $s\ge8$ .

Now
$a^2+b^2=s^2-2p=s^2-4s.$
Rather than merely substituting the smallest possible $s$ , factor the desired difference:
$a^2+b^2-32=s^2-4s-32=(s-8)(s+4)\ge0.$
Equality requires $s=8$ .
Then $(a-b)^2=s^2-8s=0$ , so $a=b=4$ ; this pair satisfies the reciprocal condition.

For part (b), $s^2-4s=45$ factors as
$(s-9)(s+5)=0.$
Because $s>0$ , only $s=9$ is allowed, and $p=18$ .
Thus $a,b$ are the roots of
$x^2-9x+18=(x-3)(x-6)=0.$
Both ordered pairs $(3,6)$ and $(6,3)$ meet the two original conditions, completing the classification.

Answer

(a) Equality only at $(a,b)=(4,4)$ . (b) $(3,6)$ and $(6,3)$ .

Key idea

Use a sum and product to connect a reciprocal equation, a square bound, and a reconstruction.
Problem 9: Mark complete
Problem 10 Three linked squares

Difficulty: 3 of 3 stars, Deep challenge

Find every triple of positive real numbers $(x,y,z)$ satisfying
$x^2+y=12,\qquad y^2+z=12,\qquad z^2+x=12.$
Prove that no unequal positive triple can work.

Builds on Factoring Quadratics, Inequality Basics
1. Hint 1
  If $x>y$ , what does comparing the first two equations tell you about $y$ and $z$ ? Continue around the three equations.
2. Hint 2
  For positive numbers, squaring preserves strict order. Show that either assumed strict order between $x$ and $y$ contradicts itself after three comparisons.
Answer

$(x,y,z)=(3,3,3)$ only.

Full solution

The equations can be written as $x^2=12-y$ , $y^2=12-z$ , and $z^2=12-x$ .
Suppose first that $x>y$ .
Positivity gives $x^2>y^2$ , so the first two rewritten equations imply $y<z$ .
Then $y^2<z^2$ , so the last two imply $z>x$ .
Finally, $z^2>x^2$ and the third and first equations imply $x<y$ , a contradiction.

If instead $x<y$ , the same comparison chain reverses all inequalities: $x^2<y^2$ gives $y>z$ ; $y^2>z^2$ gives $z<x$ ; and $z^2<x^2$ gives $x>y$ .
This is also a contradiction.
Therefore $x=y$ .
The first two equations now force $y=z$ .

Writing the common value as $t$ , we get $t^2+t=12$ , so
$(t+4)(t-3)=0.$
Positivity excludes $-4$ , leaving $t=3$ .
Direct substitution verifies the triple.
The positivity assumption is essential to the order argument: squaring arbitrary real numbers does not preserve their order.

Answer

$(x,y,z)=(3,3,3)$ only.

Key idea

In a cyclic system, follow the consequences of a strict inequality until it returns to its starting point.
Problem 10: Mark complete