If $\sqrt{x-4\sqrt{x-4}}+\sqrt{x+5-6\sqrt{x-4}}=a$, find all $a$ such that the equation has two real, distinct solutions

Question

Although I know how to start, at a certain point I get stuck. Here's what I tried: $$\sqrt{x-4\sqrt{x-4}}+\sqrt{x+5-6\sqrt{x-4}}=a \iff \sqrt{x-4 + 4\sqrt{x-4} + 4}+\sqrt{x-4-6\sqrt{x-4} + 9}=a \iff |\sqrt{x-4}-2|+|\sqrt{x-4}-3|=a$$ As well, $x\geq 4$.

1.$\sqrt{x-4}<2 \iff x - 4<4 \iff x < 8$: $$5-a = 2\sqrt{x-4} \iff 25-10+a^2 = 4(x-4)$$ And, because in the initial equality the term on the right hand is a square root, thus a positive number, $a\leq5$.

2.$2\leq \sqrt{x-4} < 3 \iff 8\leq x < 13$: $$\sqrt{x-4} - 2 + 3 - \sqrt{x-4} = a \iff a = 1$$

3.$\sqrt{x-4}\geq 3 \iff x\geq 13$: $$2\sqrt{x-4} = 5+ a \iff 4(x-4)=25 + 10a + a^2$$

From here, I am not sure how to proceed to find $a$. As well, I do not understand what the second branch is supposed to tell, that for any $x$ between $8$ and $13$, for the equality to hold, $a$ must be $1$? Any help or explanations are highly appreciated!

Answer 1

You have a good start. Now it's time to consider how many solutions for $x$ there are for given values of $a$.

1. $\sqrt{x-4} < 2$ and $5-a = 2\sqrt{x-4}$

This has one solution for $x$ if $1 < a \leq 5$.

2. $2 \leq \sqrt{x-4} < 3$ and $a=1$

If $a=1$, there are infinitely many solutions: Every value in the range $8 \leq x < 13$ satisfies the equation.

3. $\sqrt{x-4} \geq 3$ and $2\sqrt{x-4} = 5+a$

This has one solution for $x$ if $a \geq 1$.

Finally, to have exactly two solutions we must have $a \neq 1$ and a solution from case 1 and a solution from case 3. We know the value of $x$ from case 1 is not the same as the value of $x$ from case 3 since both $\sqrt{x-4} < 2$ and $\sqrt{x-4} \geq 3$ can't be true for the same value $x$. This happens if and only if $1 < a \leq 5$.

Answer 2

Substitute $\sqrt{x-4} = y$ and similarly $x = y^2+4$ and then , the equation becomes $\sqrt{y^2-4y+4} + \sqrt{y^2-6y+9} = a$

$y-2 + y-3 = a$ or, $2y-5 = a$ or, $y = \frac{a+5}{2}$

Remember what was $y$

$y = \frac{a+5}{2} = \sqrt{x-4}$

$x-4=\frac{(a+5)^2}{4}$

$x = \frac{a^2+10a+41}{4}$

$a^2 + 10a + 41 - 4x = 0$

$a = -5±\sqrt{4x-16}$

Answer 3

After $y = \sqrt{x-4}$ we have

$$ \sqrt{(y-2)^2}+\sqrt{(y-3)^2}=a $$

or

$$ |y-2|+|y-3| = a $$

and this equation has two solutions as far as $a \gt 1$ keeping in mind that $y \ge 0$ hence $1\lt a\le 2+3$

Answer 4

If

$$\sqrt{x-4\sqrt{x-4}}+\sqrt{x+5-6\sqrt{x-4}}=a, \tag1 $$ find all $a$ such that (1) above has two real, distinct solutions.

Alternative (much less elegant) approach:

I noticed that the problem was hand-crafted so that each of the two expressions, under the radicals, could be nicely re-formatted, by completing the square. This begs the question: what do you do if the problem is not hand-crafted? That is, what generic approach can you take that does not require that the expressions under the radicals allow such convenient re-formatting, via completing the square.

Throughout this response, I will tag some of the constraints with a tag formatted like (1), and some of the constraints with a tag formatted like (A.1). I will use the (A.1) format to represent Auxiliary Constraints, and refer to all of these auxiliary constraints with the syntax (A.*).

---

From (1) above, since the LHS is non-negative, you have that

$$a \geq 0. \tag{A.1}$$

Set

$$y = \sqrt{x-4} \implies y \geq 0. \tag{A.2}$$

Then, you are seeking all values of $~a~$ such that the following equation has two distinct real solutions:

$$\sqrt{y^2 + 4 - 4y} + \sqrt{y^2 + 9 - 6y} = a. \tag2 $$

So, any value of $~x~$ will satisfy (1) above if and only if (A.*) and (2) are all satisfied.

Here, I will intentionally avoid focusing on the fact that in (2) above, the expressions under the radicals each represent the square of first degree polynomials. So, I am intentionally taking a less elegant approach.

Note:
Often, in a situation like this, I will employ the strategy of Step-1 implies Step-2 implies Step-3 implies $~\cdots~$ . This results in candidate values that then have to be manually checked. In this instance, that would make the Math even more messy. So, I will instead strive to employ the strategy of Step-1 if and only if Step-2 if and only if Step-3 if and only if $~\cdots.$

In general, if you have a value $~s~$ that must be $~\geq 0,~$ and you have the equation $~s = f(x),~$ then you can conclude that the equation will be satisfied if and only if $~f(x) \geq 0 ~$ and $~s^2 = \left[~f(x)~\right]^2.$

Squaring both sides of (2) above gives:

$$(2y^2 + 13 - 10y) + 2\sqrt{(y^2 + 4 - 4y) \times (y^2 + 9 - 6y)} = a^2 \iff $$

$$2\sqrt{(y^2 + 4 - 4y) \times (y^2 + 9 - 6y)} = a^2 - (2y^2 + 13 - 10y). \tag3 $$

So, any value of $~x~$ will satisfy (1) above if and only if (A.*) and (3) are all satisfied.

(3) above implies that

$$a^2 - (2y^2 + 13 - 10y) \geq 0. \tag{A.3}$$

Squaring both sides of (3) implies that

$$4 \times (y^2 + 4 - 4y) \times (y^2 + 9 - 6y) = \left[ ~a^2 - (2y^2 + 13 - 10y) ~\right]^2. \tag{4} $$

So, any value of $~x~$ will satisfy (1) above if and only if (A.*) and (4) are all satisfied.

---

At this point, I think that it is a good idea to take a step back, and see whether this approach is viable. Fourth degree equations, by their very nature are ugly to attack. Third degree equations, although less onerous than fourth degree equations, are also no walk in the park.

However, there are three points to consider here:

• Superficial examination of the equation in (4) above indicates that both the LHS and RHS will result in a $~4y^4~$ term, which cancels out. Therefore, at worst, you will be left with a third degree polynomial.

• Often, in a situation like this, if the problem was originally hand-crafted to admit a more elegant approach, then the inelegant exploration will reveal the underlying pattern in the data.

• This entire (inelegant) approach is not a commitment to one method of attack rather than another. Instead, it is merely an exploration of the data, to see what pattern might surface.

---

Multiplying equation (4) out gives:

$$4y^4 - 40y^3 + 148y^2 - 240y + 144 \\ = 4y^4 - 40y^3 + y^2(100 + 52 - 4a^2) + y(-260 + 20a^2) + (169 - 26a^2 + a^4) \\ \implies $$

$$y^2(4 - 4a^2) + y(-20 + 20a^2) + (25 - 26a^2 + a^4) = 0. \tag{5} $$

Therefore, equation (1) above will be satisfied if and only if (A.*) and (5) are all satisfied.

---

Since $~(a^2 - 1) \times (a^2 - 25) = 25 - 26a^2 + a^4,~$
examination of (5) above indicates that the entire equation can be factored by $~(a^2 - 1).~$ This isn't really that surprising, since there would be little educational value in forcing the problem solver to grapple, from this point forward, with ugly Mathematics.

So, factoring both sides of (5) above by $~(a^2 - 1)~$ you have that (5) above will be satisfied if and only if the following equation is satisfied:

$$(a^2 - 1) \times \left[ ~y^2(-4) + y(20) + (a^2 - 25) ~\right] = 0 \iff $$

$$(a^2 - 1) \times \left[ ~4y^2 -20y + (25 - a^2) ~\right] = 0. \tag{6} $$

Therefore, (1) above will be satisfied if and only if (A.*) and (6) are all satisfied.

Discussion of $~(a^2 - 1) = 0 ~$ will be deferred to the Addendum, at the end of this answer. So, the remainder of the analysis, before the Addendum, will assume, without loss of generality, that $~a^2 \neq 1.$

This implies that

$$4y^2 -20y + (25 - a^2) = 0. \tag7 $$

(7) above will be satisfied if and only if

$$\sqrt{x-4} = y = \frac{20 \pm \sqrt{400 - 400 + 16a^2} ~}{8} = \frac{5 \pm a}{2}. \tag8 $$

Therefore, (1) above will be satisfied if and only if (A.*) and (8) are all satisfied.

As a result of (8) and (A.1) above, and the requirement that (1) above has exactly two distinct roots, you have that

$$0 \leq \frac{5 - a}{2} < \frac{5 + a}{2} \implies a \neq 0. \tag{A.4} $$

Consolidating all of the constraints against the variable $~a~$ :

• $0 < a \leq 5, ~$ so that the two roots of $~y~$ are distinct and non-negative

• $a \neq 1.$

• $x \geq 4, ~y = \sqrt{x-4} \geq 0,~$ and $~a^2 - (2y^2 + 13 - 10y) \geq 0 \implies $
  $\displaystyle (a^2 + 1) \geq 2y^2 - 10y + 14 \implies 2a^2 + 2 \geq 4y^2 - 20y + 28 \implies $
  $\displaystyle (2a^2 + 2) \geq (2y - 5)^2 + 3 \implies (2a^2 - 1) \geq (2y-5)^2.$

However, if you examine the roots of $~y~$ expressed in (8) above, you have that $~(2y-5)^2 = a^2.$

Therefore, you must have that $(2a^2 - 1) \geq a^2 \implies a^2 \geq 1.$

So, under the assumption that $~a \neq 1,~$ you have that $~a~$ must satisfy $~1 < a \leq 5.$

---

Addendum

Suppose that $~a^2 = 1.$

This implies that (6) above is satisfied.

This implies that a given value of $~x~$ satisfies (1) if and only if each of the following Auxiliary constraints are satisfied:

• (A.1) $~a \geq 0.$

• (A.2) $~\sqrt{x - 4} = y \geq 0.$

• (A.3) $~a^2 - (2y^2 + 13 - 10y) \geq 0.$

From the above constraints, and the assumption that $~a^2 = 1, ~$ you can conclude that the last bullet point above will be satisfied if and only if

$$1 \geq 2y^2 - 10y + 13 \iff 0 \geq 2y^2 - 10y + 12 \iff $$

$$0 \geq 4y^2 - 20y + 24 \iff 0 \geq (2y - 5)^2 - 1 \iff $$

$$1 \geq |2y-5| \iff 2 \leq y \leq 3 \iff $$

$$2 \leq \sqrt{x-4} \leq 3 \iff 4 \leq x-4 \leq 9 \iff 8 \leq x \leq 13.$$

So, with $~a < 0~$ rejected, since $~a~$ must be non-negative, you have that when $~a = 1,~$ then every value of $~x~$ such that $~8 \leq x \leq 13,~$ will be a solution.

Therefore, the value of $~a = 1,~$ must be rejected.