# 2006 iTest Problems/Problem U1

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## Problem

Find the real number $x$ such that $$\sqrt{x-9} + \sqrt{x-6} = \sqrt{x-1}.$$

## Solution

First, square both sides to remove some radicals. $$x-9 + 2\sqrt{(x-9)(x-6)} + x-6 = x-1$$ Next, combine terms, isolate the radical part, and square both sides. \begin{align*} x-9 + 2\sqrt{(x-9)(x-6)} + x-6 &= x-1 \\ 2x - 15 + 2\sqrt{(x-9)(x-6)} &= x-1 \\ 2\sqrt{(x-9)(x-6)} &= -x+14 \\ 4(x^2 - 15x + 54) &= x^2 - 28x + 196 \end{align*} Solve for $x$ by combining terms into a quadratic. \begin{align*} 4x^2 - 60x + 216 &= x^2 - 28x + 196 \\ 3x^2 - 32x + 20 &= 0 \\ (3x - 2)(x - 10) &= 0 \\ x &= \tfrac23 , 10 \end{align*} Because we squared both sides, we have to check for extraneous solutions. Of the two values of $x$, $\tfrac23$ does not result in equality while $10$ does. Thus, the real number $x$ that satisfies the equation is $\boxed{10}$.

## See Also

 2006 iTest (Problems) Preceded by:Problem 40 Followed by:Problem U2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
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