Question

solving radical equations classwork 3 (mixed roots)
directions: solve the following radical expressions. show your work or receive no credit!!!
$3\sqrt{x} + 4 = 13$

$3\sqrt3{x} + 4 = 13$

$4\sqrt{x} + 1 = 17$

$4\sqrt3{x} + 3 = 15$

$5\sqrt3{x} + 1 = 16$

Explanation:

Step1: Solve \( 3\sqrt{x}+4 = 13 \)

Subtract 4 from both sides: \( 3\sqrt{x}=13 - 4=9 \)
Divide both sides by 3: \( \sqrt{x}=\frac{9}{3} = 3 \)
Square both sides: \( x = 3^{2}=9 \)

Step2: Solve \( 3\sqrt[3]{x}+4 = 13 \)

Subtract 4: \( 3\sqrt[3]{x}=13 - 4 = 9 \)
Divide by 3: \( \sqrt[3]{x}=\frac{9}{3}=3 \)
Cube both sides: \( x = 3^{3}=27 \)

Step3: Solve \( 4\sqrt{x}+1 = 17 \)

Subtract 1: \( 4\sqrt{x}=17 - 1 = 16 \)
Divide by 4: \( \sqrt{x}=\frac{16}{4}=4 \)
Square: \( x = 4^{2}=16 \)

Step4: Solve \( 4\sqrt[3]{x}+3 = 15 \)

Subtract 3: \( 4\sqrt[3]{x}=15 - 3 = 12 \)
Divide by 4: \( \sqrt[3]{x}=\frac{12}{4}=3 \)
Cube: \( x = 3^{3}=27 \)

Step5: Solve \( 5\sqrt[3]{x}+1 = 16 \)

Subtract 1: \( 5\sqrt[3]{x}=16 - 1 = 15 \)
Divide by 5: \( \sqrt[3]{x}=\frac{15}{5}=3 \)
Cube: \( x = 3^{3}=27 \)

Answer:

For \( 3\sqrt{x}+4 = 13 \), \( x = 9 \);
For \( 3\sqrt[3]{x}+4 = 13 \), \( x = 27 \);
For \( 4\sqrt{x}+1 = 17 \), \( x = 16 \);
For \( 4\sqrt[3]{x}+3 = 15 \), \( x = 27 \);
For \( 5\sqrt[3]{x}+1 = 16 \), \( x = 27 \)