Question Find the solution for $x+2\sqrt{x}-8=0$

Studdy Solution

STEP 1

Assumptions
1. We are given the equation $x + 2\sqrt{x} - 8 = 0$ .
2. We are looking for the real number solutions to this equation.

STEP 2

To solve the equation, we will first try to isolate the square root term. We can do this by moving all other terms to the opposite side of the equation.
$x + 2\sqrt{x} - 8 = 0$
$2\sqrt{x} = 8 - x$

STEP 3

Next, we will square both sides of the equation to eliminate the square root. This will introduce a quadratic term.
$\left(2\sqrt{x}\right)^2 = (8 - x)^2$

STEP 4

Simplify both sides of the equation by performing the squaring operation.
$4x = (8 - x)^2$

STEP 5

Expand the right side of the equation using the binomial square formula $(a - b)^2 = a^2 - 2ab + b^2$ .
$4x = 64 - 16x + x^2$

STEP 6

Rearrange the equation to form a standard quadratic equation $ax^2 + bx + c = 0$ by moving all terms to one side.
$x^2 - 16x - 4x + 64 = 0$

STEP 7

Combine like terms on the left side of the equation.
$x^2 - 20x + 64 = 0$

STEP 8

Now, we will attempt to factor the quadratic equation. We are looking for two numbers that multiply to $64$ and add up to $-20$ .

STEP 9

The numbers that satisfy these conditions are $-16$ and $-4$ because $(-16) \times (-4) = 64$ and $(-16) + (-4) = -20$ .

STEP 10

Use these two numbers to factor the quadratic equation.
$x^2 - 16x - 4x + 64 = (x - 16)(x - 4) = 0$

STEP 11

Now, we have the factored form of the quadratic equation. We can use the zero product property to find the solutions for $x$ .
$x - 16 = 0 \quad \text{or} \quad x - 4 = 0$

STEP 12

Solve each equation for $x$ .
For $x - 16 = 0$ :
$x = 16$
For $x - 4 = 0$ :
$x = 4$

STEP 13

We have found two potential solutions, $x = 16$ and $x = 4$ . However, we must check these solutions in the original equation because squaring both sides of an equation can introduce extraneous solutions.

STEP 14

Check $x = 16$ in the original equation $x + 2\sqrt{x} - 8 = 0$ .
$16 + 2\sqrt{16} - 8 = 16 + 2 \cdot 4 - 8 = 16 + 8 - 8 = 16$
This does not satisfy the original equation, so $x = 16$ is an extraneous solution.

STEP 15

Check $x = 4$ in the original equation $x + 2\sqrt{x} - 8 = 0$ .
$4 + 2\sqrt{4} - 8 = 4 + 2 \cdot 2 - 8 = 4 + 4 - 8 = 0$
This satisfies the original equation, so $x = 4$ is a valid solution.

STEP 16

Therefore, the solution to the equation $x + 2\sqrt{x} - 8 = 0$ is $x = 4$ .

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