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Radicals and Rational Exponents: 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 Reversing a rational exponent

Difficulty: 1 of 3 stars, Stretch

Let $m,n$ be relatively prime positive integers. For real inputs, define $x^{m/n}=(\sqrt[n]{x})^m$ : when $n$ is even the root is the nonnegative real root and requires $x\ge0$ ; when $n$ is odd it is the unique real root. Use the same convention for the reduced exponent $n/m$ .

For exactly which pairs $(m,n)$ does $(x^{m/n})^{n/m}=x$ hold for every real $x$ for which the inner power is defined? Prove your classification. When it fails, give the correct formula and its real domain.

Builds on Rational Exponents
1. Hint 1
  Write $x=t^n$ using the permitted real $n$ th root, and examine the $m$ th root of $t^m$ .
2. Hint 2
  The $m$ th root of $t^m$ is $t$ for odd $m$ , but $|t|$ for even $m$ . Relative primality rules out both $m,n$ being even.
Answer

The identity holds exactly when $m$ is odd. Its domain is all real $x$ if $n$ is odd, and $x\ge0$ if $n$ is even. If $m$ is even, then $n$ is odd and the composition equals $|x|$ on all of $\mathbb R$ .

Full solution

Let $t=\sqrt[n]{x}$ according to the stated convention.
Then $x=t^n$ and
$x^{m/n}=t^m.$
If $m$ is odd, the unique real $m$ th root of $t^m$ is $t$ .
Raising that result to the power $n$ gives $t^n=x$ .
The outer power is defined for every permitted inner input.
The inner domain is all real numbers when $n$ is odd and the nonnegative real numbers when $n$ is even.

If $m$ is even, relative primality forces $n$ to be odd.
Thus the inner power is defined for every real $x$ , and $t$ can have either sign.
The nonnegative $m$ th root of $t^m$ is $|t|$ , so the outer operation gives $|t|^n=|x|$ .
In particular, at any negative $x$ it differs from $x$ , disproving the proposed identity for these pairs.

This exhausts the parity cases.
It also explains the failure of simply multiplying the two rational exponents: an even principal root discards a sign that a later power cannot recover.
The conventions and the domain are part of the identity, not optional annotations.

Answer

The identity holds exactly when $m$ is odd. Its domain is all real $x$ if $n$ is odd, and $x\ge0$ if $n$ is even. If $m$ is even, then $n$ is odd and the composition equals $|x|$ on all of $\mathbb R$ .

Key idea

Exponent multiplication must respect the real-root convention, especially when an even principal root appears.
Problem 1: Mark complete
Problem 2 Spacing between nearby square roots

Difficulty: 1 of 3 stars, Stretch

For a real number $x>1$ , define $D(x)=2\sqrt x-\sqrt{x-1}-\sqrt{x+1}$ . Prove that $D(x)>0$ , and explain what this says about the gaps between the three square roots.

By rationalizing twice, prove the quantitative bounds
$\frac1{4(x+1)^{3/2}}<D(x)<\frac1{4(x-1)^{3/2}}.$
1. Hint 1
  Write $D(x)$ as $(\sqrt x-\sqrt{x-1})-(\sqrt{x+1}-\sqrt x)$ and rationalize each difference.
2. Hint 2
  After combining the two fractions, rationalize the new numerator $\sqrt{x+1}-\sqrt{x-1}$ . Bound each of the three resulting denominator factors between $2\sqrt{x-1}$ and $2\sqrt{x+1}$ .
Answer

$D(x)>0$ , so $\sqrt x-\sqrt{x-1}>\sqrt{x+1}-\sqrt x$ . The two strict bounds hold for every $x>1$ .

Full solution

All square roots and the proposed bounding denominators are positive because $x>1$ .
Rationalizing the two successive differences gives $D(x)=1/(\sqrt x+\sqrt{x-1})-1/(\sqrt{x+1}+\sqrt x)$ .
The second denominator is larger than the first, so $D(x)>0$ .
Thus moving one unit to the right produces a smaller increase in the square root.

Combining these fractions leaves numerator $\sqrt{x+1}-\sqrt{x-1}$ .
Rationalizing it gives the exact identity $D(x)=2/((\sqrt{x+1}+\sqrt{x-1})(\sqrt x+\sqrt{x-1})(\sqrt{x+1}+\sqrt x))$ .
Every factor in this denominator is strictly larger than $2\sqrt{x-1}$ and strictly smaller than $2\sqrt{x+1}$ .

The product of the three positive factors is therefore strictly between $8(x-1)^{3/2}$ and $8(x+1)^{3/2}$ .
Taking reciprocals reverses the inequalities, and multiplying by $2$ yields exactly the claimed bounds.
These estimates quantify the shrinking gap using only conjugates and order comparisons; no approximate square-root values are needed.

Answer

$D(x)>0$ , so $\sqrt x-\sqrt{x-1}>\sqrt{x+1}-\sqrt x$ . The two strict bounds hold for every $x>1$ .

Key idea

Rationalizing more than once can expose both the sign and the size of a small difference of radicals.
Problem 2: Mark complete
Problem 3 An inequality with two radical sides

Difficulty: 1 of 3 stars, Stretch

Find all real $x$ for which
$\sqrt{x+4}+\sqrt{x-1}\ge\sqrt{5x}.$
Identify every equality case. In your proof, state why each squaring step preserves equivalence, rather than just producing a necessary condition.
1. Hint 1
  First determine the common domain. Both sides are then nonnegative, so the first squaring is reversible.
2. Hint 2
  After the first squaring, isolate the remaining radical. On the domain, the other side is $3x-3\ge0$ , so a second squaring is also reversible.
Answer

The solution set is $[1,5]$ , with equality at $x=1$ and $x=5$ .

Full solution

The three square roots are defined together exactly when $x\ge1$ .
On this domain both sides of the original inequality are nonnegative.
Squaring therefore preserves their order and gives $2x+3+2\sqrt{(x+4)(x-1)}\ge5x$ , or
$2\sqrt{x^2+3x-4}\ge3x-3.$

The left side is nonnegative, and the right side is nonnegative because $x\ge1$ .
Consequently, squaring this inequality is again reversible.
It becomes
$4(x^2+3x-4)\ge9(x-1)^2.$
Simplifying gives $-5x^2+30x-25\ge0$ , equivalently
$(x-1)(x-5)\le0.$
Its real solutions are exactly $1\le x\le5$ , all within the original domain.

Because every step was an equivalence on that domain, the interval is both necessary and sufficient; no extraneous candidates remain.
Equality in the last factored inequality occurs at $1$ and $5$ .
Substitution confirms $\sqrt5+0=\sqrt5$ at the first, and $3+2=5$ at the second.
The sign check before each squaring is what allows a whole interval to be recovered safely.

Answer

The solution set is $[1,5]$ , with equality at $x=1$ and $x=5$ .

Key idea

Squaring an inequality is reversible only after the signs of both sides are controlled.
Problem 3: Mark complete
Problem 4 A nested radical that levels off

Difficulty: 2 of 3 stars, Challenge

For real $x$ for which it is defined, let
$F(x)=\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}.$
Find the exact real domain and a piecewise formula for $F$ . For every real $c$ , find all solutions of $F(x)=c$ . Explain why one output has infinitely many preimages.
1. Hint 1
  Set $t=\sqrt{x-1}\ge0$ , so $x=t^2+1$ . Both outer radicands become perfect squares.
2. Hint 2
  The square roots simplify to $t+1$ and $|t-1|$ . Separate $0\le t\le1$ from $t\ge1$ .
Answer

Domain $[1,\infty)$ ; $F(x)=2\sqrt{x-1}$ for $1\le x\le2$ , and $F(x)=2$ for $x\ge2$ . For $0\le c<2$ : $x=1+c^2/4$ ; for $c=2$ : all $x\ge2$ ; otherwise no solutions.

Full solution

The inner radical requires $x\ge1$ .
Set $t=\sqrt{x-1}$ , which is nonnegative, and write $x=t^2+1$ .
The two outer radicands become $t^2+1+2t=(t+1)^2$ and
$t^2+1-2t=(t-1)^2.$
They are nonnegative for every $t\ge0$ , so no further domain restriction is needed.

Principal square roots give
$F=t+1-|t-1|.$
If $0\le t\le1$ , this equals $2t$ ; if $t\ge1$ , it equals $2$ .
Translating to $x$ gives the stated piecewise formula, with both branches agreeing at $x=2$ .
Omitting the absolute value would incorrectly make $F$ equal to $2$ even at $x=1$ , where its actual value is $0$ .

Thus no output below $0$ or above $2$ is possible.
For $0\le c<2$ , only the first branch can apply, and $2\sqrt{x-1}=c$ gives the unique permitted input $x=1+c^2/4$ .
For $c=2$ , every $x\ge2$ works, including the shared endpoint.
The infinite preimage occurs because beyond $t=1$ both outer roots increase by exactly the same amount, leaving their difference constant.

Answer

Domain $[1,\infty)$ ; $F(x)=2\sqrt{x-1}$ for $1\le x\le2$ , and $F(x)=2$ for $x\ge2$ . For $0\le c<2$ : $x=1+c^2/4$ ; for $c=2$ : all $x\ge2$ ; otherwise no solutions.

Key idea

Denesting a radical may reveal an absolute value and a constant branch, not a single algebraic formula.
Problem 4: Mark complete
Problem 5 A shortest three-stage path

Difficulty: 2 of 3 stars, Challenge

Real numbers $x,y$ satisfy $0\le x\le y\le8$ . Find the minimum possible value of
$L=\sqrt{x^2+4}+\sqrt{(y-x)^2+9}+\sqrt{(8-y)^2+25},$
and every pair $(x,y)$ attaining it. Prove your result without calculus. You may interpret the terms as lengths of consecutive line segments whose vertical rises are $2,3,5$ .

Schematic; $x$ and $y$ are variable.

Text description of this figure

A horizontal axis labeled X and a vertical axis labeled Y meet at the point A. A path of three straight segments, with a dot at each corner, climbs from A to a point B, from B to a point C, and from C to a point D. B is at height 2 above the horizontal position x, C is at height 5 above the horizontal position y, and D is at height 10 above the horizontal position 8. Dashed guide lines run from B, C and D across to the Y axis, where the heights 2, 5 and 10 are marked, and down to the X axis, where the positions x, y and 8 are marked. The drawing is schematic, and x and y are variable.
1. Hint 1
  Place consecutive vertices at $(0,0),(x,2),(y,5),(8,10)$ . The expression is the length of the resulting broken path.
2. Hint 2
  Compare this path to the straight segment from $(0,0)$ to $(8,10)$ . Equality requires the three positive vertical rises to have the same horizontal-to-vertical ratio.
Answer

Minimum $2\sqrt{41}$ , attained only at $(x,y)=(8/5,4)$ .

Full solution

Let $A=(0,0)$ , $B=(x,2)$ , $C=(y,5)$ , and $D=(8,10)$ .
By the distance formula,
$L=AB+BC+CD.$
The straight segment joining the endpoints has length
$AD=\sqrt{8^2+10^2}=2\sqrt{41}.$
A broken path cannot be shorter than the straight segment, so
$L\ge2\sqrt{41}.$

An algebraic justification of the two-segment inequality uses horizontal and vertical displacements $(p,q)$ and $(r,s)$ .
The identity $(p^2+q^2)(r^2+s^2)-(pr+qs)^2=(ps-qr)^2\ge0$ implies
$pr+qs\le\sqrt{p^2+q^2}\sqrt{r^2+s^2}.$
Adding $p^2+q^2+r^2+s^2$ to twice this inequality and taking nonnegative square roots shows that the length of the combined displacement is at most the sum of the two lengths.
Applying this twice proves the bound.

Equality requires all three segments to point along the same straight line.
Since their vertical rises are positive, their horizontal displacements must be in the ratio $2:3:5$ .
Their total horizontal displacement is $8$ , so $x=8(2/10)=8/5$ and
$y=8((2+3)/10)=4.$
This pair obeys $0\le x\le y\le8$ and makes all four vertices collinear in order, so equality is attained.
The fixed rises force these displacements uniquely.

Answer

Minimum $2\sqrt{41}$ , attained only at $(x,y)=(8/5,4)$ .

Key idea

A sum of square roots may encode a broken path whose shortest configuration is forced by collinearity.
Problem 5: Mark complete
Problem 6 A system with negative cube roots

Difficulty: 2 of 3 stars, Challenge

Find all ordered pairs of real numbers $(x,y)$ satisfying
$x+y=7,\qquad x^{2/3}+y^{2/3}=5.$
Here $x^{2/3}=(\sqrt[3]x)^2$ and cube roots are real. Your solution must allow negative inputs and justify the rejection of every other algebraic candidate.
1. Hint 1
  Set $u=\sqrt[3]x,v=\sqrt[3]y$ and then $s=u+v$ . The square-sum condition gives both $uv=(s^2-5)/2$ and $s^2\le10$ .
2. Hint 2
  Use $u^3+v^3=s^3-3uvs$ to obtain $(s-1)(s^2+s-14)=0$ . Check the bound $s^2\le10$ before recovering $u,v$ .
Answer

Exactly $(x,y)=(8,-1)$ and $(-1,8)$ .

Full solution

Let $u=\sqrt[3]x,v=\sqrt[3]y$ , both allowed to be negative.
Then $u^2+v^2=5$ and $u^3+v^3=7$ .
Put $s=u+v$ .
We have $uv=(s^2-5)/2$ , while $(u-v)^2\ge0$ gives
$s^2\le2(u^2+v^2)=10.$

The cube identity yields $7=s^3-3uvs=(15s-s^3)/2$ , or
$s^3-15s+14=(s-1)(s^2+s-14)=0.$
Besides $s=1$ , the algebraic candidates are $r_+=(-1+\sqrt{57})/2$ and
$r_-=(-1-\sqrt{57})/2.$
Since $7<\sqrt{57}<9$ , we have $3<r_+<4$ and $r_-<-4$ .
The negative candidate has square greater than $16$ , violating $s^2\le10$ .
The positive candidate satisfies $r_+^2=14-r_+>10$ , also violating it.

Thus $s=1$ is the only feasible sum, giving $uv=-2$ .
The numbers $u,v$ are the zeros of $t^2-t-2=(t-2)(t+1)$ , so they are $2,-1$ in either order.
Cubing gives the two listed pairs.
Directly, $8+(-1)=7$ and
$8^{2/3}+(-1)^{2/3}=4+1=5.$
These checks confirm that the negative input is essential and valid.

Answer

Exactly $(x,y)=(8,-1)$ and $(-1,8)$ .

Key idea

A substitution may create algebraic candidates that violate the bounds inherited from real variables.
Problem 6: Mark complete
Problem 7 An inverse hidden in two cube roots

Difficulty: 2 of 3 stars, Challenge

For every real $t$ , define
$C(t)=\sqrt[3]{t+\sqrt{t^2+1}}+\sqrt[3]{t-\sqrt{t^2+1}},$
using real cube roots. Prove that $C$ is strictly increasing and takes every real value exactly once. Find an explicit formula for its inverse function, and evaluate $C(18)$ without decimal approximations.

Builds on Inverse Functions
1. Hint 1
  Call the two cube-root terms $u,v$ . Their product is particularly simple, even though one term is negative.
2. Hint 2
  If $s=C(t)$ , expand $(u+v)^3$ . To prove uniqueness, compare $r^3+3r$ and $s^3+3s$ by factoring their difference.
Answer

$C^{-1}(s)=(s^3+3s)/2$ for every real $s$ , and $C(18)=3$ .

Full solution

The square root is defined for every real $t$ , and both real cube roots are defined even though
$t-\sqrt{t^2+1}<0.$
Let their values be $u,v$ .
Then $u^3+v^3=2t$ and $(uv)^3=(t+\sqrt{t^2+1})(t-\sqrt{t^2+1})=-1$ , so $uv=-1$ .
For $s=u+v$ , the cube identity gives $s^3=2t-3s$ , or $s^3+3s=2t$ .

The function $H(s)=s^3+3s$ is strictly increasing without needing derivatives: for $r>s$ , $H(r)-H(s)=(r-s)(r^2+rs+s^2+3)>0$ , because
$r^2+rs+s^2=(r+s/2)^2+3s^2/4\ge0.$
Consequently, the identity $H(C(t))=2t$ implies that $C$ is strictly increasing.

To prove that every output is attained, choose any real $s$ and put
$t=(s^3+3s)/2.$
The already-defined value $w=C(t)$ satisfies
$H(w)=2t=H(s).$
Strict increase of $H$ forces $w=s$ .
Thus the proposed inverse works for every real input and each output occurs once.
In particular, $H(3)=27+9=36=2\cdot18$ , so $C(18)=3$ .

Answer

$C^{-1}(s)=(s^3+3s)/2$ for every real $s$ , and $C(18)=3$ .

Key idea

An intimidating radical formula may be understood by deriving the much simpler equation it inverts.
Problem 7: Mark complete
Problem 8 Sharp bounds for three coupled radicals

Difficulty: 3 of 3 stars, Deep challenge

Nonnegative real numbers $a,b,c$ satisfy $a+b+c=1$ . Find the least and greatest possible values of
$R=\sqrt{a(1-b)}+\sqrt{b(1-c)}+\sqrt{c(1-a)}.$
Give every equality case and prove both bounds. If you use an inequality involving sums of products, derive the needed version from nonnegative squares.
1. Hint 1
  For the lower bound, $1-b=a+c\ge a$ , with two analogous comparisons.
2. Hint 2
  For the upper bound, use $u=(\sqrt a,\sqrt b,\sqrt c)$ and $v=(\sqrt{1-b},\sqrt{1-c},\sqrt{1-a})$ . Expand $(\sum u_i^2)(\sum v_i^2)-(\sum u_iv_i)^2$ into three squares.
Answer

$1\le R\le\sqrt2$ . The minimum occurs exactly at $(1,0,0),(0,1,0),(0,0,1)$ ; the maximum exactly at $(1/3,1/3,1/3)$ .

Full solution

The constraint implies $0\le a,b,c\le1$ , so every radical is real.
Since $a(1-b)=a(a+c)\ge a^2$ , its nonnegative square root is at least $a$ .
The analogous bounds give
$R\ge a+b+c=1.$
Equality requires $ac=ab=bc=0$ , so at most one variable is positive.
The sum constraint then gives exactly the three listed vertices, each of which attains $1$ .

For the upper bound, let $u_1=\sqrt a,u_2=\sqrt b,u_3=\sqrt c$ and
$v_1=\sqrt{1-b},v_2=\sqrt{1-c},v_3=\sqrt{1-a}.$
Expanding shows $(\sum u_i^2)(\sum v_i^2)-(\sum u_iv_i)^2=(u_1v_2-u_2v_1)^2+(u_1v_3-u_3v_1)^2+(u_2v_3-u_3v_2)^2\ge0$ .
The two squared sums are $1$ and $2$ , so $R^2\le2$ , hence
$R\le\sqrt2.$

Equality above requires the two nonzero vectors to be proportional.
Their squared lengths force $v_i=\sqrt2u_i$ , giving $1-b=2a$ , $1-c=2b$ , and $1-a=2c$ .
Solving yields $a=b=c=1/3$ .
Conversely, at this triple each term is $\sqrt2/3$ , giving $R=\sqrt2$ .
This also proves uniqueness of the maximum, including all possible zero-variable boundary cases.

Answer

$1\le R\le\sqrt2$ . The minimum occurs exactly at $(1,0,0),(0,1,0),(0,0,1)$ ; the maximum exactly at $(1/3,1/3,1/3)$ .

Key idea

Choose a representation that makes both a square-based bound and its equality conditions transparent.
Problem 8: Mark complete
Problem 9 How much degree does a radical require?

Difficulty: 3 of 3 stars, Deep challenge

Let $\alpha=\sqrt2+\sqrt5$ . Find the monic polynomial with rational coefficients of smallest possible degree having $\alpha$ as a zero. Prove that no nonzero rational-coefficient polynomial of smaller degree can vanish at $\alpha$ , and list all the zeros of your polynomial.

For the degree proof, first show that $\sqrt5$ cannot equal $u+v\sqrt2$ for rational $u,v$ . You may use the elementary fact that the prime exponents in a nonzero rational square are even.
1. Hint 1
  Squaring $\alpha$ isolates a single square root; square once more to construct a quartic.
2. Hint 2
  Prove that $1,\sqrt2,\sqrt5,\sqrt{10}$ have no nonzero rational linear relation by grouping a relation as $A+B\sqrt2=-(C+D\sqrt2)\sqrt5$ . Then expand a general cubic expression in $\alpha$ .
Answer

The unique monic polynomial of smallest degree is $x^4-14x^2+9$ , of degree $4$ . Its four distinct zeros are $\sqrt2+\sqrt5,\sqrt2-\sqrt5,-\sqrt2+\sqrt5,-\sqrt2-\sqrt5$ .

Full solution

Since $\alpha^2=7+2\sqrt{10}$ , squaring $\alpha^2-7=2\sqrt{10}$ gives
$\alpha^4-14\alpha^2+9=0.$
The quartic factors as $(x^2-(7+2\sqrt{10}))(x^2-(7-2\sqrt{10}))$ , giving the four listed distinct real zeros.

For minimality, suppose $\sqrt5=u+v\sqrt2$ with rational $u,v$ .
Squaring gives $5=u^2+2v^2+2uv\sqrt2$ , so irrationality of $\sqrt2$ forces $uv=0$ .
Either $u^2=5$ or $v^2=5/2$ would follow, impossible for a rational square by prime exponents.
Now suppose $A+B\sqrt2+C\sqrt5+D\sqrt{10}=0$ with rational coefficients.
If $C+D\sqrt2\ne0$ , division and rationalization would express $\sqrt5$ as $u+v\sqrt2$ , just ruled out.
The rationalizing denominator $C^2-2D^2$ cannot vanish unless $C=D=0$ .
Therefore $C=D=0$ , and then irrationality gives $A=B=0$ .

Finally,
$\alpha^3=17\sqrt2+11\sqrt5.$
A relation $a\alpha^3+b\alpha^2+c\alpha+d=0$ becomes
$(7b+d)+(17a+c)\sqrt2+(11a+c)\sqrt5+2b\sqrt{10}=0.$
The independence just proved forces $b=d=0$ and $17a+c=11a+c=0$ , hence $a=c=0$ .
So no nonzero polynomial of degree at most $3$ works.
The quartic is minimal; another monic quartic with the same zero would differ from it by such a polynomial of degree at most $3$ , proving uniqueness.

Answer

The unique monic polynomial of smallest degree is $x^4-14x^2+9$ , of degree $4$ . Its four distinct zeros are $\sqrt2+\sqrt5,\sqrt2-\sqrt5,-\sqrt2+\sqrt5,-\sqrt2-\sqrt5$ .

Key idea

Constructing an equation gives an upper bound on algebraic degree; proving minimality requires ruling out every lower-degree relation.
Problem 9: Mark complete
Problem 10 A nested equation with a changing number of solutions

Difficulty: 3 of 3 stars, Deep challenge

For every real parameter $a\ge0$ , find the exact real domain of the expression $\sqrt{a-\sqrt{a-x}}$ and every real solution of
$\sqrt{a-\sqrt{a-x}}=x.$
Give a complete classification of the number of distinct solutions. Identify which solutions also satisfy $\sqrt{a-x}=x$ , and explain the extra solutions when they occur.
1. Hint 1
  Set $y=\sqrt{a-x}$ . At a solution, both $x,y$ are nonnegative and satisfy $x^2+y=a$ and $y^2+x=a$ .
2. Hint 2
  Subtract these equations to obtain $(x-y)(x+y-1)=0$ . In the unequal case, impose nonnegativity on both roots of $t^2-t+1-a=0$ .
Answer

Domain $[a-a^2,a]$ . There is always the solution $r=(\sqrt{1+4a}-1)/2$ . For $3/4<a\le1$ , there are also $x=(1\pm\sqrt{4a-3})/2$ , giving three distinct solutions; otherwise $r$ is the only one. Only $r$ solves $\sqrt{a-x}=x$ .

Full solution

The inner radical requires $x\le a$ .
The outer radical requires $\sqrt{a-x}\le a$ ; because $a\ge0$ , this is equivalent to
$a-x\le a^2.$
Thus the expression has domain
$a-a^2\le x\le a.$
At a solution, set
$y=\sqrt{a-x}\ge0.$
The outer equality also gives $x\ge0$ , and squaring the two nonnegative equalities is equivalent to $y^2+x=a$ and $x^2+y=a$ .

Subtraction gives
$(x-y)(x+y-1)=0.$
If $x=y$ , then $x^2+x=a$ , whose unique nonnegative root is
$r=(\sqrt{1+4a}-1)/2.$
If $x\ne y$ , then $x+y=1$ , so both numbers are roots of
$t^2-t+1-a=0.$
They are real when $a\ge3/4$ , nonnegative when $a\le1$ , and distinct when $a>3/4$ .
This produces the two additional formulas precisely for $3/4<a\le1$ .
At $a=3/4$ they coincide at $1/2=r$ , adding no new solution.

Conversely, the fixed pair $(r,r)$ and every stated unequal pair have nonnegative entries and satisfy both squared equations.
Taking principal square roots recovers $\sqrt{a-x}=y$ and $\sqrt{a-y}=x$ , so all candidates satisfy the original nested equation and its domain.
An unequal pair is swapped by one application of $x\mapsto\sqrt{a-x}$ and returns after two; it is not a fixed point.
Thus there are exactly three solutions in the specified interval of parameters and one in every other case.

Answer

Domain $[a-a^2,a]$ . There is always the solution $r=(\sqrt{1+4a}-1)/2$ . For $3/4<a\le1$ , there are also $x=(1\pm\sqrt{4a-3})/2$ , giving three distinct solutions; otherwise $r$ is the only one. Only $r$ solves $\sqrt{a-x}=x$ .

Key idea

Turn a nested radical into a symmetric system, then check the sign conditions that make squaring reversible.
Problem 10: Mark complete