Math  /  Algebra

Question EQUAZIONE 1 GRADO (2(31)12x2^\frac{\text { EQUAZIONE } 1 \text { GRADO }}{(\sqrt[(3-1)^{1} 2-x \hat{2}]{2}}

Studdy Solution

STEP 1

What is this asking? We're asked to solve the equation (2x1)2=x2(\sqrt{2x-1})^2 = x^2 for xx. Watch out! Squaring both sides of an equation can introduce extraneous solutions, so we'll need to check our answers!

STEP 2

1. Simplify the equation
2. Solve for xx
3. Check for extraneous solutions

STEP 3

Let's **get rid** of that pesky square root!
We know that squaring a square root simplifies things nicely.
Since (a)2=a(\sqrt{a})^2 = a, we can simplify the left side of our equation: (2x1)2=2x1(\sqrt{2x-1})^2 = 2x-1.
So our equation becomes 2x1=x22x - 1 = x^2.
Much better!

STEP 4

To solve for xx, let's **rewrite** our equation in standard quadratic form: ax2+bx+c=0ax^2 + bx + c = 0.
We can do this by subtracting 2x2x and adding 1 to both sides of 2x1=x22x - 1 = x^2.
This gives us x22x+1=0x^2 - 2x + 1 = 0.

STEP 5

Now, we can **factor** this quadratic equation.
We're looking for two numbers that multiply to **1** and add up to **-2**.
Those numbers are -1 and -1!
So, we can rewrite our equation as (x1)(x1)=0(x-1)(x-1) = 0, or even more compactly as (x1)2=0(x-1)^2 = 0.

STEP 6

To find the solutions, we set each factor equal to zero.
In this case, we have x1=0x - 1 = 0.
Adding 1 to both sides gives us x=1x = 1.

STEP 7

It's super important to **check** if our solution works in the *original* equation to avoid any extraneous solutions.
Let's plug x=1x = 1 back into (2x1)2=x2(\sqrt{2x-1})^2 = x^2: (211)2=12(\sqrt{2 \cdot 1 - 1})^2 = 1^2 (21)2=1(\sqrt{2 - 1})^2 = 1(1)2=1(\sqrt{1})^2 = 11=11 = 1

STEP 8

Since 1=11 = 1, our solution x=1x = 1 is valid!

STEP 9

The solution to the equation (2x1)2=x2(\sqrt{2x-1})^2 = x^2 is x=1x = 1.

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