2006 iTest Problems/Problem U1

Revision as of 01:36, 1 December 2018 by Rockmanex3 (talk | contribs) (Solution to Problem U1 -- Radical Equation Combo)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
Invalid username
Login to AoPS