Question

solving a radical equation in exercises 27–38, solve the equation, if possible. check your solutions. 27. \\(\sqrt{5x} - 10 = 0\\) 28. \\(\sqrt{3x + 1} = 7\\) 29. \\(4 + \sqrt3{2x - 9} = 0\\)

Explanation:

Problem 27: $\boldsymbol{\sqrt{5x} - 10 = 0}$

Step1: Isolate the radical term

Add 10 to both sides of the equation:
$\sqrt{5x} = 10$

Step2: Eliminate the square root

Square both sides to remove the square root:
$(\sqrt{5x})^2 = 10^2$
$5x = 100$

Step3: Solve for $x$

Divide both sides by 5:
$x = \frac{100}{5} = 20$

Step4: Check the solution

Substitute $x = 20$ back into the original equation:
Left-hand side: $\sqrt{5(20)} - 10 = \sqrt{100} - 10 = 10 - 10 = 0$
Right-hand side: $0$
Since both sides are equal, $x = 20$ is valid.

Problem 28: $\boldsymbol{\sqrt{3x + 1} = 7}$

Step1: Eliminate the square root

Square both sides to remove the square root:
$(\sqrt{3x + 1})^2 = 7^2$
$3x + 1 = 49$

Step2: Solve for $x$

Subtract 1 from both sides:
$3x = 49 - 1 = 48$
Divide by 3:
$x = \frac{48}{3} = 16$

Step3: Check the solution

Substitute $x = 16$ back into the original equation:
Left-hand side: $\sqrt{3(16) + 1} = \sqrt{48 + 1} = \sqrt{49} = 7$
Right-hand side: $7$
Since both sides are equal, $x = 16$ is valid.

Problem 29: $\boldsymbol{4 + \sqrt[3]{2x - 9} = 0}$

Step1: Isolate the cube root term

Subtract 4 from both sides:
$\sqrt[3]{2x - 9} = -4$

Step2: Eliminate the cube root

Cube both sides to remove the cube root:
$(\sqrt[3]{2x - 9})^3 = (-4)^3$
$2x - 9 = -64$

Step3: Solve for $x$

Add 9 to both sides:
$2x = -64 + 9 = -55$
Divide by 2:
$x = \frac{-55}{2} = -27.5$

Step4: Check the solution

Substitute $x = -27.5$ back into the original equation:
Left-hand side: $4 + \sqrt[3]{2(-27.5) - 9} = 4 + \sqrt[3]{-55 - 9} = 4 + \sqrt[3]{-64} = 4 + (-4) = 0$
Right-hand side: $0$
Since both sides are equal, $x = -27.5$ (or $\frac{-55}{2}$) is valid.

Answer:

s:

1. $\boldsymbol{x = 20}$
2. $\boldsymbol{x = 16}$
3. $\boldsymbol{x = -\frac{55}{2}}$ (or $\boldsymbol{x = -27.5}$)