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Karolina [17]
3 years ago
14

} " alt=" x - 2\sqrt{x - 3 = 0} " align="absmiddle" class="latex-formula">
what value of x satisfies the equation above?
x=​
Mathematics
1 answer:
GrogVix [38]3 years ago
7 0

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.

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