A Sum of Roots

For getting a real solution none of the equation under the square root should be less than zero after substituting the solution, take second and third square root terms from the equation, by equating it to zero, we get x=40.5, and for x<40.5 or x>40.5, in both cases one of the term gets negative under the square root which gives a conjugate number. So, just check the equation by substituting the value of x in the equation, the total gets equal to zero, and it is the only solution by which this equation gets satisfied, if you take any other value, one of the term will gets into a conjugate number. Therefore the solution is 40.5. So, as given a/b= 40.5, given a and b are positive coprime numbers, 81 and 2 are coprime numbers (which are mutually prime) satisfying a/b =40.5, then we get a=81, b=2, by adding it a+b = 81+2 =83. Therefore, the solution is 83

well first we need to observe the equation...specially the second and last expressions. and the solution has to be real, so in the square root it has to be a positive or zero in both cases. there is no chance of being positive so since one is the negative of the other. the remaining option is being zero. we compute the expression in the square root and it gives x=81/2 positive co-prime numbers . a=81 and b=2 , hence a+b=83

√(4x² + 26x – 129) – √(81 – 2x) – √(4x² – 2x + 1734 + √2x – 81) = 0

√(4x² + 26x – 129) = √(81 – 2x) + √(4x² – 20 + 1734 + √2x – 81)

√(4x² + 26x – 129) = √(4x² – 20 + 1734

(4x² + 26x – 129) = (4x² – 20 + 1734

46x = 1863

2x= 81

x = 81/2

81 + 2 = 83

The problem does not say whether there can be any negative numbers underneath the square roots. Here is a summary of a complete solution considering negative numbers under the square roots as a possibility. First manipulate the LHS of the equation into $\sqrt{4x^2+26x-129}-\sqrt{4x^2-20x+1734}+\sqrt{2x-81}-i\sqrt{2x-81}=0$ .We know that the result of each square root will give either a real number or a pure imaginary number. Since it is true that if $\sqrt{2x-81}$ is a real number, then $-i\sqrt{2x-81}$ is a pure imaginary number and vice-versa, if one assumes that either both $\sqrt{4x^2+26x-129}$ and $\sqrt{4x^2-20x+1734}$ give real numbers or both give pure imaginary numbers, then one ends up with $\sqrt{2x-81}=0$ (or, equivalently, $x=\frac{81}{2}$ ) by examining the imaginary part or real part of the equation respectively. When one substitutes $x=\frac{81}{2}$ into the original equation, one finds that it is a solution, and that, in actuality, $\sqrt{4x^2+26x-129}=\sqrt{4x^2-20x+1734}=\sqrt{7485}$ (so they happen to be real). The remaining possibilities can all be covered in four cases by assuming one of $\sqrt{4x^2+26x-129}$ and $\sqrt{4x^2-20x+1734}$ gives a real number and the other a pure imaginary number, whilst $\sqrt{2x-81}$ could be real or pure imaginary. Let case 1 be the case where $\sqrt{4x^2+26x-129}$ and $\sqrt{2x-81}$ are real and $\sqrt{4x^2-20x+1734}$ is pure imaginary. In that case, one gets the equations $\sqrt{4x^2+26x-129}+\sqrt{2x-81}=0$ and $-\sqrt{-4x^2+20x-1734}-\sqrt{2x-81}=0$ by equating the real and imaginary parts (on each side of the original equation). Since the square root of a nonnegative real number is nonnegative, the square roots in each equation must equal zero. At this point, whether you determined $-4x^2+20x-1734=0$ has no real roots or $2x-81=0$ produces a solution we already found (and there are in actuality no solutions), it is not hard to see that we can immediately skip over this case. Case 2 of 4: $\sqrt{4x^2-20x+1734}$ and $\sqrt{2x-81}$ are real and $\sqrt{4x^2+26x-129}$ is pure imaginary: equating the real and imaginary parts leads to the equations $\sqrt{2x-81}-\sqrt{4x^2-20x+1734}=0$ and $\sqrt{-4x^2-26x+129}-\sqrt{2x-81}=0$ . Adding the equations results in $-\sqrt{4x^2-20x+1734}+\sqrt{-4x^2-26x+129}=0\Rightarrow \sqrt{-4x^2-26x+129}=\sqrt{4x^2-20x+1734}\Rightarrow -4x^2-26x+129=4x^2-20x+1734$ , but the roots of this equation are both imaginary, so we can move to case 3. Case 3 of 4: $\sqrt{4x^2+26x-129}$ is real and $\sqrt{4x^2-20x+1734}$ and $\sqrt{2x-81}$ are pure imaginary: this leads to the equations $\sqrt{4x^2+26x-129}+\sqrt{81-2x}=0$ and $-\sqrt{-4x^2+20x-1734}+\sqrt{81-2x}=0$ . We can conclude in a similar fashion to case 1 that there is no solution to these equations. Case 4 of 4: $\sqrt{4x^2-20x+1734}$ is real and $\sqrt{4x^2+26x-129}$ and $\sqrt{2x-81}$ are pure imaginary: we can achieve reduction to the following pair of equations: $-\sqrt{4x^2-20x+1734}+\sqrt{81-2x}=0$ and $\sqrt{-4x^2-26x+129}+\sqrt{81-2x}=0$ . Again, via similar reasoning, there are no solutions in this case. Consequently, the only real solution is $x=\frac{81}{2}$ .

(81-2x)^1/2 and (2x-81)^1/2 shows that we can square root them.I don't know how to explain but we can take for an example like 7-3=4 and 3-7=-4.So we can conclude that the both term are equal to 0 because we cannot square root of negative number.

so 81-2x=2x-81

we got x =40.5=81/2

a+b=83

1. Consider the term \sqrt{81-2x} in the left side of equation. We know that since x is real number, then 81-2x \geq 0 \Rightarrow x \leq \frac {81}{2}
2. Now if we consider the term \sqrt{2x - 81} in the right side, with the same reason, we get x \leq \frac {81}{2}.
3. Thus we can conclude that x = \frac {81}{2}
4. And so the answer is 81 + 2 = 83

First of all, the solution(s) should satisfy the domain of each term in the equation. But let's just focus on the 2nd and 4th term. We can see that taking the negative of the base (the one inside the square root) in the 2nd term results to the base of the 4th term, and vice versa. Now, if we solve for the domain of the 2nd term, we will get: $x <= \frac{81}{2}$ . On the other hand, taking the domain of the 4th will result to $x >= \frac{81}{2}$ . Since they are in the same equation, they should have the same value of x. Taking the intersection of their domains, we will get $x = \frac{81}{2}$ . Therefore, $81$ + $2$ = $\boxed {83}$ . (Note: Sorry for the formatting. :P)