Question

} " alt=" x - 2\sqrt{x - 3 = 0} " align="absmiddle" class="latex-formula">
what value of x satisfies the equation above?
x=​

Answer 1

Answer:

No Real Solutions

Step-by-step explanation:

So we have:

$x-2\sqrt{x-3}=0$

First, determine the domain restrictions. The expression under the radical cannot be less than 0. Therefore:

$x-3\geq 0\\x\geq 3$

Therefore, our final answers must be greater than or equal to 3:

Now, going back to the original equation, subtract x from both sides:

$-2\sqrt{x-3}=-x$

Now, square both sides:

$4(x-3)=x^2$

Distribute:

$4x-12=x^2$

Subtract 4x and add 12 to both sides:

$0=x^2-4x+12$

This isn't factor-able. Let's use the quadratic formula. a is 1, b is -4 and c is 12:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substitute:

$x=\frac{4\pm\sqrt{(-4)^2-4(1)(12)}}{2(1)}$

Simplify the radical:

$x=\frac{4\pm\sqrt{-32}}{2(1)}$

The number under the radical is negative. In other words, there are no real solutions.