Math  /  Algebra

QuestionSolve and Verify  Solve and verify 4.c) 3x+4=x2\begin{array}{l} \text { Solve and verify } 4 \text {.c) } \sqrt{3 x+4}=\sqrt{x-2} \end{array}

Studdy Solution

STEP 1

What is this asking? We need to find the value of xx that makes the square root of 3x+43x + 4 equal to the square root of x2x - 2, and then we need to double-check our answer! Watch out! Remember, we can't take the square root of a negative number!
So, whatever answer we get for xx must make *both* 3x+43x + 4 and x2x - 2 greater than or equal to zero.

STEP 2

1. Square Both Sides
2. Isolate xx
3. Verify the Solution

STEP 3

We have an equation with square roots on both sides: 3x+4=x2\sqrt{3x + 4} = \sqrt{x - 2}.
To get rid of the square roots, we can **square both sides** of the equation!
This is totally allowed, as long as we do the *same thing* to *both* sides.

STEP 4

(3x+4)2=(x2)2 (\sqrt{3x + 4})^2 = (\sqrt{x - 2})^2 Squaring a square root removes the radical, so we get: 3x+4=x2 3x + 4 = x - 2

STEP 5

To get all the xx terms on one side, let's **subtract** xx from both sides of the equation: 3x+4x=x2x 3x + 4 - x = x - 2 - x 2x+4=2 2x + 4 = -2

STEP 6

Now, let's **subtract** 4 from both sides to isolate the term with xx: 2x+44=24 2x + 4 - 4 = -2 - 4 2x=6 2x = -6

STEP 7

Finally, to get xx all by itself, we **divide** both sides by 2: 2x2=62 \frac{2x}{2} = \frac{-6}{2} x=3 x = -3

STEP 8

Let's plug our **solution**, x=3x = -3, back into the *original* equation: 3(3)+4=32 \sqrt{3 \cdot (-3) + 4} = \sqrt{-3 - 2}

STEP 9

9+4=5 \sqrt{-9 + 4} = \sqrt{-5} 5=5 \sqrt{-5} = \sqrt{-5}

STEP 10

Uh oh!
We've run into a problem.
We can't take the square root of a negative number!
This means our solution, x=3x = -3, isn't actually a valid solution.

STEP 11

There is *no solution* to the given equation.
Even though we found x=3x = -3 algebraically, it doesn't work when we plug it back in because it leads to taking the square root of a negative number.

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