Question

Which of the following equations is/are quadratic equation(s)?

q₁: (y + 1)² = 2y

q₂: (y – 1)² = y²

q₃: (y + 1)³ = (y – 1)³

`q_4: 1 + sqrt(y) = (sqrt(y) + 1)^2`

Options

• q₁, q₂ and q₄

• q₁ and q₂

• q₁, q₃ and q₄

• q₁ and q₃

Answer 1

q₁ and q₃

Explanation:

1. Simplify q₁: (y + 1)² = 2y

Expanding the left side: y² + 2y + 1 = 2y

Subtracting 2y from both sides: y² + 1 = 0

This is a quadratic equation because the highest power of the variable is 2.

2. Simplify q₂: (y – 1)² = y²

Expanding the left side: y² – 2y + 1 = y²

Subtracting y² from both sides: –2y + 1 = 0

This is a linear equation, not quadratic, because the y² terms cancel out.

3. Simplify q₃: (y + 1)³ = (y – 1)³

Expanding both sides using (a ± b)³ = a³ ± 3a²b + 3ab² ± b³:

(y³ + 3y² + 3y + 1) = (y³ – 3y² + 3y – 1)

The y³ and 3y terms cancel: 3y² + 1 = –3y² – 1 

Combining terms: 6y² + 2 = 0

This is a quadratic equation because the y³ terms cancel, leaving y² as the highest power.

4. Simplify q₄: `1 + sqrt(y) = (sqrt(y) + 1)^2`

Expanding the right side: `1 + sqrt(y) = (sqrt(y))^2 + 2sqrt(y) + 1`

Simplifying: `1 + sqrt(y) = y + 2sqrt(y) + 1`

Combining terms: `y + sqrt(y) = 0`

This is not a quadratic equation because it contains a square root term `(sqrt(y))`, meaning it is not a polynomial in y.