Exactly 3 Real Solutions Allowed

Main idea: By manipulating the given equation, we can obtain two quadratic equations. Either one of them has a double root, or they share a common root. Try each case.

Complete proof:

Let $n = x^2+4x+136$ , because the column is too narrow otherwise.

We first observe the following:

$(x^2 + 4x + 4a + 136)^2 = 16a (5x^2 + 4x + 136)$

$(n+4a)^2 = 16a(n+4x^2)$

$n^2 + 2n(4a) + (4a)^2 = 16an + 16a(4x^2)$

$n^2 + 8na + (4a)^2 = 16na + 64ax^2$

$n^2 - 8na + (4a)^2 = 64ax^2$

$n^2 - 2(n)(4a) + (4a)^2 = (8x\sqrt{a})^2$

$(n-4a)^2 = (8x\sqrt{a})^2$

$(x^2+4x+136-4a)^2 = (8x\sqrt{a})^2$

$x^2+4x+136-4a = 8x\sqrt{a}$ or $x^2+4x+136-4a = -8x\sqrt{a}$

Let $u = \sqrt{a}$ ; remember that $u \ge 0$ .

$x^2+4x+136-4u^2 = 8xu$ or $x^2+4x+136-4u^2 = -8xu$

$x^2 + (4-8u)x + (136-4u^2) = 0$ or $x^2 + (4+8u)x + (136-4u^2) = 0$

Now, we must obtain exactly three real solutions. So either one of them has double roots or one root is shared among both equations.

• Case 1: Some of them has a double root.

Suppose that the first equation has a double root. Then its discriminant must be zero; that is, $(4-8u)^2 - 4(136-4u^2) = 0$ . Then,

$16 - 64u + 64u^2 - 544 + 16u^2 = 0$

$80u^2 - 64u - 528 = 0$

$16(5u^2 - 4u - 33) = 0$

So this gives a nonnegative solution $u = 3$ . (The result $u = -\frac{11}{5}$ can be thrown away as $u$ is nonnegative.)

The same method can be used to the second equation to obtain $16(5u^2 + 4u - 33) = 0$ , which gives a nonnegative solution $u = \frac{11}{5}$ .

Now we need to check each of our results.

For the first result, our equations become $x^2 - 20x + 100 = 0$ and $x^2 + 28x + 100 = 0$ . The first equation gives the double root $x = 10$ and the second equation gives the roots $x = \dfrac{-28 \pm \sqrt{28^2 - 4 \cdot 100}}{2} = -14 \pm 4\sqrt{6}$ (by abc formula) which are distinct from each other and from $10$ , so this solution, $u = 3$ or $a = u^2 = 9$ , works.

For the second result however, $u = \frac{11}{5}$ gives a negative discriminant for the first equation:

$(4-8u)^2 - 4(136-4u^2)$

$= \left(4 - 8 \cdot \frac{11}{5} \right)^2 - 4 \left( 136 - 4 \left( \dfrac{11}{5} \right)^2 \right)$

$= \left( \dfrac{-68}{5} \right)^2 - 4 \left( 136 - \dfrac{484}{25} \right)$

$= \dfrac{4624}{25} - \dfrac{11664}{25}$

$= \dfrac{-1408}{5} < 0$

So the first equation has no real roots. So this result fails.

• Case 2: The two equations share a common root.

So $x^2 + (4-8u)x + (136-4u^2) = 0$ and $x^2 + (4+8u)x + (136-4u^2) = 0$ . Subtracting the two equations together gives $-16ux = 0$ . Either $u = 0$ or $x = 0$ .

When $u = 0$ , we have $x^2 + 4x + 136 = 0$ from the first equation, which doesn't have any real root. So this result fails.

When $x = 0$ , we have $136-4u^2 = 0$ , or $u = \sqrt{34}$ , or $a = 34$ . Checking, the first equation becomes $x^2 + (4-8\sqrt{34})x = 0$ and the second equation becomes $x^2 + (4+8\sqrt{34})x = 0$ . After removing the common root $x = 0$ , the first equation gives $x = 8\sqrt{34} - 4$ and the second $x = -8\sqrt{34} - 4$ , so we have exactly three real roots here.

So in total, $a = 9$ and $a = 34$ are the only ones that work, and so the sum is $9+34 = \boxed{43}$ .

We know that $\forall m,n: (m+n)^2-(m-n)^2=4mn$

Because the LHS of the equation looks like $(m+n)^2$ and the RHS looks like $4mn$ , we try to take the difference of the following squares $\ (x^2+4x+4a+136)^2-(x^2+4x-4a+136)^2$

$=4.4a(x^2+4x+136)$

$=16a(x^2+4x+136)$ , which only differs from the RHS of the equation by $64ax^2$

The given equation is now equivalent to $(x^2+4x-4a+136)^2=64ax^2$ $\Leftrightarrow x^2+4x-4a+136=8 \sqrt{a}x$ (1)
or $x^2+4x-4a+136=-8 \sqrt{a}x$ (2) (since a is positive as given)

In order for the given equation to have 3 distinct real roots, either

• one of the equations (1) and (2) have 2 real roots and the other have 1 repeated root, of which all 3 are distinct; or

• both equations (1) and (2) have 2 real roots but they have a common root.

Case 1: One of the equations (1) and (2) have 2 roots and the other have 1 root

Let's examine the discriminants of equation 1 and 2.

$\Delta_1 = (4-8\sqrt{a})^2-4(136-4a)=16(-33+5a-4\sqrt{a})\geq0$

$\Leftrightarrow (\sqrt{a}-3)(5\sqrt{a}+11)\geq0$ $\Leftrightarrow \sqrt{a}\geq3$ or $a\geq9$ (since $\sqrt{a}\geq0$ )

Similarly, $\Delta_2\geq0 \Leftrightarrow (\sqrt{a}+3)(5\sqrt{a}-11)\geq0$ $\Leftrightarrow \sqrt{a}\geq \frac{11}{5}$ or $a\geq\frac{121}{25}$ (since $\sqrt{a}\geq0$ )

Since $9>\frac{121}{25}$ , $a=9$ will give equation (1) 1 repeated root and equation (2) 2 distinct real roots.

By plugging $a=9$ into (1) and (2), we can easily verify that the given quartic equation has indeed 3 distinct real solutions.

Case 2: both equations (1) and (2) have 2 real roots but they have a common root.

Let $x_0$ be the common root of both (1) and (2) then $8 \sqrt{a}x_0=-8\sqrt{a}x_0$ $\Leftrightarrow a=0$ or $x_0=0$

$a=0$ does not satisfy the conditions of discriminants therefore it is rejected.

Substitute $x=0$ into the equations to get $a=34$ . We easily verify $x=0$ as the common solution and as $a>9>\frac{121}{25}$ , both equations have 2 real roots.

The answer is therefore $34+9=43$

(x² + 4x + 4a + 136)² = 16a(5x² + 4x + 136)

{(x² + 4x - 4a + 136) + 8a}² = 16a(5x² + 4x + 136)

(x² + 4x - 4a + 136)² + 16a(x² + 4x - 4a + 136) + 64a² = 16a(5x² + 4x + 136)

(x² + 4x - 4a + 136)² - 64ax² = 0

{x² - (8√a - 4)x - (4a - 136)}{x² + (8√a + 4)x - (4a - 136)} = 0

Let a = b²,

{x² - (8b - 4)x - (4b² - 136)}{x² + (8b + 4)x - (4b² - 136)} = 0 .... ❶

We can see the equation ❶ is a product of two polynomials. Assume the first polynomial is

f₁(x) = x² - (8b - 4)x - (4b² - 136)

and the second polynomial is

f₂(x) = x² + (8b + 4)x - (4b² - 136)

Denote the discriminant of f₁(x) is Δ₁ and the discriminant of f₂(x) is Δ₂.

Δ₁ = (8b - 4)² + 4(4b² - 136)

Δ₁ = 80b² - 64b - 528

Δ₂ = (8b + 4)² + 4(4b² - 136)

Δ₂ = 80b² + 64b - 528

Since a ≥ 0, b ≥ 0, Δ₁ ≤ Δ₂, we may divide into two cases.

First case:

Δ₁ > 0, Δ₂ > 0, with one same root.

Second case:

Δ₁ = 0, Δ₂ > 0

Hence, we get a = 43.

Expanding the given terms and collecting like terms, we obtain a monic quartic polynomial. $(x^2+4x+4a+136)^2=16a(5x^2+4x+136)$ $(x^2+4x+4a+136)^2-16a(5x^2+4x+136)=0$ $(x^2+4x+4a+136)(x^2+4x+4a+136)-(80ax^2+64ax+16(136)a)=0$ $(x^2)(x^2+4x+4a+136)+(4x)(x^2+4x+4a+136)+(4a)(x^2+4x+4a+136)+136(x^2+4x+4a+136)-(80ax^2+64ax+16(136)a)=0$ $(x^4+4x^3+4ax^2+136x^2+4x^3+16x^2+16ax+4(136)x+4ax^2+16ax+16a^2+4(136)a+136x^2+(136)4x+(136)4a+(136)^2-80ax^2-64ax-16(136)a)=0$ $(x^4+4x^3+4ax^2+136x^2+4x^3+16x^2+16ax+4(136)x+4ax^2+16ax+16a^2+4(136)a+136x^2+(136)4x+(136)4a+(136)^2-80ax^2-64ax-16(136)a)=0$ $x^4+8x^3+288x^2-72ax^2+1088x-32ax+18496-1088a+16a^2=0$ We will name this polynomial $f(x)$ . Clearly, as the coefficients of this polynomial are real (as well as integral, please not) and as $a$ is also real, the quartic cannot have a non-real complex root, or else its conjugate must also be a root of $f$ , violating the three real solution parameter addressed in the problem. Thus, the only viable solution type of the quartic is a real double root and two other distinct real roots. For this to occur, the discriminant of the polynomial must be zero. From the definition for the quartic discriminant found on Wolfram MathWorld (under the page “Polynomial Discriminants”), the discriminant for the above quartic function is $337,929,100,591,104a^2-127,245,922,336,768a^3+14,365,860,298,752a^4-549,218,942,976a^5+6,710,886,400a^6$ Yikes. We’re supposed to find the roots of this giant sextic polynomial mess. Thankfully, it is easy to notice that an $a^2$ can be pulled out of the expression, leading to the hunt of only four zeroes. Note that the resultant constant term, $337,929,100,591,104$ is divisible by 9 (the sum of the digits is 54) and that this term is also divisible by 11. By the rational root theorem, these two numbers just may be roots of the polynomial. Trying out if $a-9$ is a factor by synthetic division, we find that it is indeed. Just because we have a gut feeling that $a-34$ is a factor of the expression, we try it out. Indeed, it works, and it works again, so we have successfully factored the polynomial as $a^2(a-9)(a-34)^2(25a-121)$ Now the possible real values of $a$ are $0,9,34$ , and $\frac{121}{25}$ . Of these, we will show that $\frac{121}{25}$ is not a viable value for $a$ as it produces only one distinct real root. This is easy to prove. If we substitute $\frac{121}{25}$ into the equation for $f(x)$ , we obtain $x^4+8x^3+288x^2-72\frac{121}{25}x^2+1088x-32\frac{121}{25}x+18496-1088\frac{121}{25}+16\frac{121}{25}^2=0$ Using some division, we find that the above is equal to $\frac{1}{625}(5x+54)^2(25x^2-340x+2916)=0$ , with only non-real complex-valued factors of the trinomial expression. Thus, we know that the only viable solutions for $a$ are $0,9,$ and $34$ , whose sum is $\boxed{43}$ .

Let $b = x^2 + 4x + 136$ . Then the given equation can be rewritten as:

$(1) (b + 4a)^2 = 16a(4x^2) + 16ab$

After transferring $16ab$ to the left hand side of the equation, $(1)$ can be written thus:

$(2) (b - 4a)^2 = (64x^2)a$

Taking the square root of both sides, we have two equations, which can be expressed in terms of $x$ . ( $b$ is now replaced by $x^2 + 4x + 136$ .)

$(3i) x^2 + (4 - 8\sqrt{a})x + (136 - 4a) = 0$ $(3ii) x^2 + (4 + 8\sqrt{a})x + (136 - 4a) = 0$

Using the quadratic formula to solve for $x$ ,

$(4i) x = 4\sqrt{a} - 2 \pm 2\sqrt{5a - 4\sqrt{a} - 33}$ $(4ii) x = -4\sqrt{a} - 2 \pm 2\sqrt{5a + 4\sqrt{a} - 33}$

We can now consider three cases for all possible values of $a$ and $x$ :

CASE 1: $5a - 4\sqrt{a} - 33 = 0$

If such is the case for certain values of $\sqrt{a}$ , then from $(4i)$ , there will be one value for $x$ . Also, $\sqrt{5a + 4\sqrt{a} - 33}$ , which is greater than $\sqrt{5a - 4\sqrt{a} - 33}$ because $\sqrt{a} \geq{0}$ , will give two distinct values for $x$ , for a total of three real solutions for $x$ .

Solving for $\sqrt{a}$ , we get two possible values, one of which $(\sqrt{a} = -\frac{11} {5})$ is rejected due to the condition for $\sqrt{a}$ . We only take $\sqrt{a} = 3$ , or $a = 9$ . With this value of $a$ in $(4i)$ , we get $x = 10$ ; from $(4ii)$ , we get $x = -14 \pm 2\sqrt{6}$ . We found exactly three real solutions for $x$ in the original equation when $a = 9$ , which makes $9$ one of the required values for $a$ .

(A similar solution can be used for $\sqrt{5a + 4\sqrt{a} - 33} = 0$ . In this case, only $\sqrt{a} = \frac {11} {5}$ , or $a = \frac{121} {25}$ , is accepted as a possible value for $a$ . However, for this value of $a$ , $\sqrt{5a - 4\sqrt{a} - 33}$ gives two complex roots. Hence, $\frac{121} {25}$ does not satisfy the conditions of the problem.)

CASE 2: $5a - 4\sqrt{a} - 33 > 0$

In this case, there are two possible distinct values for $x$ in $(4i)$ . Also, $5a + 4\sqrt{a} - 33 > 0$ , which when plugged into $(4ii)$ will give another two distinct solutions for a total of four values of $x$ . Hence, no values of $a$ under Case 2 satisfy the conditions of the problem.

CASE 3: $x$ = 0. (We'll do this case for $(4i)$ ; a similar solution can be done for $(4ii)$ .)

Since $\sqrt{5a - 4\sqrt{a} - 33} \geq{0}$ , which only happens when $a \geq{9}$ , $(4a) \geq{0}$ . Taking the equality part and solving for $a$ , we get $a = 34$ . Substituting this value of $a$ in $(4ii)$ , we get, after simplification of the nested radical, $x = -4\sqrt{34} - 2 \pm (2\sqrt{34} + 1)$ - which gives three real solutions of $x$ for $a = 34$ . (In the other similar case, from $(4ii)$ ), we could show that $x = 0$ when $a = 34$ .)

Since $9$ and $34$ are the only positive values of $a$ that satisfy the given problem, the required answer is $9 + 34 = 43$ .

If \­(4a=b>0\­) then the equation is equivalent to : $(x^{2}+4x+136-b)^{2}=16bx^{2}$ We have these two cases: $x^{2}+4x+136-b=4sqrt(b)x$ or $x^{2}+4x+136-b=-4sqrt(b)x$ We search for double roots of the first and the second equation, having in mind that if one of them has a double root, the other one must have 2 distinct roots. We find that $\sqrt{b}=6$ or $\sqrt{b}=22/5$ but only $\sqrt{b}=6$ fulfils the desired conditions. Then we search for roots of the first and the second equation which coincide. Because $\sqrt{b}>0$ , we easily find that $\sqrt{b}=2\sqrt{34}$ . So if $b=36$ then $a=9$ and if $b=4*34$ then $a=34$ and the answer is $34+9=\boxed{43}$

$(x^{2}+4x+136)^{2}=16a(5x^{2}+4x+136)$ $(x^{2}+4x+136)^{2}+16a^{2}+8a(x^{2}+4x+136)=64ax^{2}+16a(x^{2}+4x+136)$ $(x^{2}+4x+136-4a)^{2}=64ax^{2}$ $(x^{2}+4x+136-4a)^{2}-(8x\sqrt{a})^{2}=0$ $(x^{2}+(4+8\sqrt{a})x+136-4a)(x^{2}+(4-8\sqrt{a})x+136-4a)=0$ To get 3 real solution we can get it with play the discriminant. the probability ,for a value of "a" ,one equation have a same root and other equation has two root.

If Discriminant from $(x^{2}+(4+8\sqrt{a})x+136-4a)=0$ we know that only have 1 solution for x in that equation. from D=0 we get $\sqrt{a}=\frac{11}{5}$ and $\sqrt{a}=-3$ or $a=\frac{121}{25}$ and $a=9$

subtitution the value of a to $(x^{2}+(4-8\sqrt{a})x+136-4a)$ for $\sqrt{a}=\frac{11}{5}-->$ x has no real solution so $a=\frac{121}{25}$ not satisfy for $\sqrt{a}=-3-->$ x has two real solution so a=9 is satisfy

and last form, $(x^{2}+(4+8\sqrt{a})x+136-4a)(x^{2}+(4-8\sqrt{a})x+136-4a)=0$ two equation has one same root we get a=0 or x=0, for a=0 has no real root for x if x=0 we can get 136-4a = 0 so a=34 is satisfy too.

and then the sum is 34+9 = 43

On expanding the LHS and then bunching terms we get

x^4 + 8x^3 + x^2(288 - 72a) + x (1088 - 32a) + (4a + 136)^2 - (136)*16a= ‡ 0

Now we apply the condition that for a Quartic Equation to have three real solutions, D2<0. D1 and D2 are as follows: D1=2b^3-9abc+27da^2 D2=D1^2-4(b^2-3ac)^3

Applying D2<0 on our quartic equation we get a=9 and a=34 (These are the values for a in which the +/- root part in the solutions for x becomes zero)

Given -: (x^2+4x+4a+136)^2=16a(5x^2+4x+136) (x^2+4x+4a+136)^2=16a(136+4x+5x^2) (x^2+4x+4a+136)^2=16(136a+4ax+5ax^2) (x^2+4x+4a+136)^2=80ax^2+64ax+2176a (x^2+4x+4a+136)*(x^2+4x+4a+136)=80ax^2+64ax+2176a on expanding LHS terms: 16 a^2+8 a x^2+32 a x+1088 a+x^4+8 x^3+288 x^2+1088 x+18496 = 80 a x^2+64 a x+2176 a Solving the 4th degree equation above: x = 2 (2 \sqrt{a}-\sqrt{5 a-4 \sqrt{a}-33}-1) x = 2 (2 \sqrt{a}+\sqrt{5 a-4 \sqrt{a}-33}-1) x = 2 (-2 \sqrt{a}-\sqrt{5 a+4 \sqrt{a}-33}-1) x = 2 (-2 \sqrt{a}+\sqrt{5 a+4 \sqrt{a}-33}-1)

now we get three real solutions for 1). a=9 x=10, x= -14-4\sqrt{6}, x= 4\sqrt{6}-14 2). a=34 x=0, x= -4-8\sqrt{34}, x= 8\sqrt{34}-4 The answer is : 9 + 34 = 43

above equation can be written as

(x^2+4x-4a+136)^2=64ax^2

from the above we'll get the two quadratic equation as shown below

x^2+(4-8sqrt(a)x+136-4a=0 and x^2+(4+8sqrt(a)x+136-4a=0

since we have exactly three real root,so we can say that one of the root is repeated,so lets see the determinant of above two equation which are

D1=(4-8sqrt(a)^2-4(136-4a)=80a-64sqrt(a)-528 and D2=80a+64sqrt(a)-528 respectively

one of D1 and D2 must be 0 another one must be greater than 0

since a>0 D1 must be 0 and D2>0

putting D1=0 will give a=9

to get the other solution we have to constant value of above two quadratics t be zero so that we'll get the solution as 0,0,-4+8sqrt(a),-4-8sqrt(a) which are three distinct real roots roots also.

thus 4a-136=0 => a=34

thus the required answer is 9+34=43