Question

rationalizing a denominator the process for rationalizing a denominator in a variable expression is the same as in a numeric expression. here’s a real - world example. the kinetic energy of the car of a rollercoaster is given by the formula ( k=\frac{1}{2}mv^{2} ), where ( k ) is kinetic energy, ( m ) is the mass of the car, and ( v ) is the velocity of the car. solving this formula for ( v ), we get ( v = sqrt{\frac{2k}{m}} ). which formula gives the velocity of the car in simplest form?
options:
( v=\frac{sqrt{2k}}{m} )
( v=\frac{sqrt{2km}}{2m} )
( v=\frac{sqrt{2km}}{m} )

Explanation:

Step1: Start with the given formula

We have \( v = \sqrt{\frac{2k}{m}} \). To simplify this, we rationalize the denominator (or simplify the radical expression) by manipulating the square - root. Recall that \( \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} \) for \( a\geq0,b > 0 \), and to rationalize the denominator of \( \frac{\sqrt{a}}{\sqrt{b}} \), we multiply the numerator and denominator by \( \sqrt{b} \).

Step2: Simplify the radical

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Let's check the options:

• The first option is \( v=\frac{\sqrt{2km}}{m} \), the second option has a denominator of \( 2m \) which is incorrect, and the third option has \( \sqrt{2k} \) in the numerator without multiplying by \( \sqrt{m} \) to rationalize the denominator.

Answer:

The first option (the left - most formula) \( v=\frac{\sqrt{2km}}{m} \)