Question

Solve these 4 radical equations:

a) y = 3√x - 2
b) y = -5 + 2√x - 4
c) y = 3√x - 1
d) y = 3√x - 4 + 1

Answer 1

a)
`\( y = 3√(x) - 2 \)`

b)
`\( y = -5 + 2√(x) - 4 \)`

c)
`\( y = 3√(x) - 1 \)`

d) )
`\( y = 3√(x) - 4 + 1 \)`

Step-by-step explanation:

Radical equations involve solving for variables within a root (√) expression. Let's solve each equation step-by-step:

a)
`\( y = 3√(x) - 2 \):`

To solve for \(x\), isolate the radical term:
`\( 3√(x) = y + 2 \)`. Then, square both sides to eliminate the square root and solve for
`\(x\).`

b)
`\( y = -5 + 2√(x) - 4 \):`

Start by isolating the radical term:
`\( 2√(x) = y + 9 \)`. Square both sides to get rid of the radical and solve for
`\(x\).`

c)
`\( y = 3√(x) - 1 \):`

Isolate the radical term:
`\( 3√(x) = y + 1 \)`. Square both sides to eliminate the radical and solve for
`\(x\).`

d)
`\( y = 3√(x) - 4 + 1 \):`

Simplify the equation:
`\( y = 3√(x) - 3 \).` Isolate the radical term:
`\( 3√(x) = y + 3 \)`. Square both sides and solve for
`\(x\).`

Remember, while solving radical equations, it's crucial to check solutions back into the original equation to ensure they're valid. Squaring both sides can sometimes introduce extraneous solutions. Verify each solution by substituting it back into the original equation to guarantee its accuracy.

By methodically solving these equations and confirming the solutions, we can ascertain the accurate values of \(x\) and \(y\) that satisfy each equation.

Answer 2

The solutions to the given radical equations are:

`a) \( x = (1)/(27)(y + 2)^3 \)`

`b) \( x = (1)/(4)(y + 9)^2 \)`

`c) \( x = (1)/(27)(y + 1)^3 \)`

`d)`
`\( x = (1)/(27)(y - 5)^3 \)`

Step-by-step explanation:

To solve these radical equations, we need to isolate the variable
`\( x \)` in each equation.

For equation a), we can cube both sides, rearrange terms, and solve for
`\( x \):`

`\[ x = (1)/(27)(y + 2)^3 \]`

For equation b), we can square both sides, rearrange terms, and solve for
`\( x \):`

`\[ x = (1)/(4)(y + 9)^2 \]`

For equation c), we can cube both sides, rearrange terms, and solve for
`\( x \)`:

`\[ x = (1)/(27)(y + 1)^3 \]`

For equation d), we can cube both sides, rearrange terms, and solve for
`\( x \):`

`\[ x = (1)/(27)(y - 5)^3 \]`

These solutions represent the relationship between
`\( x \)` and
`\( y` for each given equation. They are obtained through algebraic manipulation of the original equations to isolate
`\( x \)` on one side. The process involves applying inverse operations to the radical terms and simplifying the expressions.