Question

enter the values for the variables that give the correct simplified expressions, ( x geq 0 ).

( sqrt{50x^2} = sqrt{25 cdot 2 cdot x^2} = 5xsqrt{b} )
( b = square )

( sqrt{32x} = sqrt{16 cdot 2 cdot x} = csqrt{2x} )
( c = square )

( sqrt{18n} = sqrt{9 cdot 2 cdot n} = esqrt{2n} )
( e = square )

( sqrt{72x^2} = sqrt{36 cdot 2 cdot x^2} = qxsqrt{2} )
( q = square )

Explanation:

Step1: Solve for \( b \)

Given \( \sqrt{50x^2} = \sqrt{25 \cdot 2 \cdot x^2} = 5x\sqrt{b} \). After simplifying \( \sqrt{25 \cdot 2 \cdot x^2} \), we get \( 5x\sqrt{2} \) (since \( \sqrt{25}=5 \), \( \sqrt{x^2}=x \) for \( x\geq0 \), and the remaining factor under the square root is 2). So comparing with \( 5x\sqrt{b} \), we see \( b = 2 \).

Step2: Solve for \( c \) (first \( c \))

For \( \sqrt{32x} = \sqrt{16 \cdot 2 \cdot x} = c\sqrt{2x} \). Simplify \( \sqrt{16 \cdot 2 \cdot x} \), \( \sqrt{16}=4 \), so we have \( 4\sqrt{2x} \). Comparing with \( c\sqrt{2x} \), we get \( c = 4 \).

Step3: Solve for \( c \) (second \( c \))

For \( \sqrt{18n} = \sqrt{9 \cdot 2 \cdot n} = c\sqrt{2n} \). Simplify \( \sqrt{9 \cdot 2 \cdot n} \), \( \sqrt{9}=3 \), so we have \( 3\sqrt{2n} \). Comparing with \( c\sqrt{2n} \), we get \( c = 3 \).

Step4: Solve for \( g \)

For \( \sqrt{72x^2} = \sqrt{36 \cdot 2 \cdot x^2} = gx\sqrt{2} \). Simplify \( \sqrt{36 \cdot 2 \cdot x^2} \), \( \sqrt{36}=6 \), \( \sqrt{x^2}=x \) for \( x\geq0 \), so we have \( 6x\sqrt{2} \). Comparing with \( gx\sqrt{2} \), we get \( g = 6 \).

Answer:

\( b = 2 \), \( c = 4 \) (for \( \sqrt{32x} \)), \( c = 3 \) (for \( \sqrt{18n} \)), \( g = 6 \)

(In the order of the blanks: first blank \( b = 2 \), second blank \( c = 4 \), third blank \( c = 3 \), fourth blank \( g = 6 \))