Math  /  Algebra

QuestionWhich statement about xx5=1\sqrt{x}-\sqrt{x-5}=1 is true? x=4x=4 is a true sivelution. x=4x=4 is an extraneous solution. x=9x=9 is a true solution. x=9x=9 is an extraneous solution.

Studdy Solution

STEP 1

What is this asking? We're given an equation with square roots, and we need to figure out if x=4x=4 and/or x=9x=9 are actual solutions or if they're just pretending to be solutions (extraneous). Watch out! When dealing with square roots, sometimes we get solutions that look good but don't actually work in the original equation.
Those sneaky solutions are called extraneous solutions!

STEP 2

1. Test x=4
2. Test x=9

STEP 3

Let's **plug in** x=4x = 4 into our equation xx5=1\sqrt{x} - \sqrt{x-5} = 1.
This gives us 445=1\sqrt{4} - \sqrt{4-5} = 1.

STEP 4

Now, 4\sqrt{4} is simply **2**.
And 454-5 is **-1**, so we have 21=12 - \sqrt{-1} = 1.

STEP 5

Uh oh!
We have the square root of a negative number!
That's not a real number, so x=4x=4 **cannot** be a solution.
It's **extraneous**!

STEP 6

Now, let's try x=9x = 9.
Substituting into xx5=1\sqrt{x} - \sqrt{x-5} = 1 gives us 995=1\sqrt{9} - \sqrt{9-5} = 1.

STEP 7

We know 9\sqrt{9} is **3**, and 959-5 is **4**, so we have 34=13 - \sqrt{4} = 1.

STEP 8

Since 4\sqrt{4} is **2**, our equation becomes 32=13 - 2 = 1.

STEP 9

Hey, 323 - 2 *does* equal **1**, so 1=11 = 1!
This means x=9x = 9 is a **true solution**!

STEP 10

x=4x = 4 is an extraneous solution, and x=9x = 9 is a true solution.

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