Problem:
If a ⩾ 1 a \geqslant 1a ⩾ 1 , then the sum of the real solutions of
a − a + x = x \sqrt{a-\sqrt{a+x}}=x
a − a + x ​ ​ = x
is equal to
Answer Choices:
A. a − 1 \sqrt{a}-1a ​ − 1
B. a − 1 2 \dfrac{\sqrt{a}-1}{2}2 a ​ − 1 ​
C. a − 1 \sqrt{a-1}a − 1 ​
D. a − 1 2 \dfrac{\sqrt{a-1}}{2}2 a − 1 ​ ​
E. 4 a − 3 − 1 2 \dfrac{\sqrt{4 a-3}-1}{2}2 4 a − 3 ​ − 1 ​
Solution:
Since x xx is the principal square root of some quantity, x ⩾ 0 x \geqslant 0x ⩾ 0 . For x ⩾ 0 x \geqslant 0x ⩾ 0 , the given equation is equivalent to
a − a + x = x 2 a-\sqrt{a+x}=x^{2}
a − a + x ​ = x 2
Since the left member of this equation is a decreasing function of x xx and the right member is an increasing function, one easily verifies that the equation has exactly one solution. To find this solution let y = a + x y=\sqrt{a+x}y = a + x ​ . Then
a − y = x 2 a − y − y 2 = x 2 − y 2 a − y − ( a + x ) = x 2 − y 2 − ( x + y ) = ( x + y ) ( x − y ) 0 = ( x + y ) ( x − y + 1 ) \begin{gathered}
a-y=x^{2} \\
a-y-y^{2}=x^{2}-y^{2} \\
a-y-(a+x)=x^{2}-y^{2} \\
-(x+y)=(x+y)(x-y) \\
0=(x+y)(x-y+1)
\end{gathered}
a − y = x 2 a − y − y 2 = x 2 − y 2 a − y − ( a + x ) = x 2 − y 2 − ( x + y ) = ( x + y ) ( x − y ) 0 = ( x + y ) ( x − y + 1 ) ​
Since a ⩾ 1 a \geqslant 1a ⩾ 1 and x ⩾ 0 x \geqslant 0x ⩾ 0 , it follows that y > 0 y>0y > 0 and x + y ≠0 x+y \neq 0x + y î€ = 0 . Therefore,
x − y + 1 = 0 x + 1 = y x + 1 = a + x ( x + 1 ) 2 = a + x x = − 1 ± 4 a − 3 2 . \begin{gathered}
x-y+1=0 \\
x+1=y \\
x+1=\sqrt{a+x} \\
(x+1)^{2}=a+x \\
x=\dfrac{-1 \pm \sqrt{4 a-3}}{2} .
\end{gathered}
x − y + 1 = 0 x + 1 = y x + 1 = a + x ​ ( x + 1 ) 2 = a + x x = 2 − 1 ± 4 a − 3 ​ ​ . ​
The positive solution x = 4 a − 3 − 1 2 x=\dfrac{\sqrt{4 a-3}-1}{2}x = 2 4 a − 3 ​ − 1 ​ is the sum.