Question

Which of the following is a quadratic equation?

• A. \( (x+1)(x-1) = x^2 - 4x^3 \)
• B. \( (x+4)^2 = 3x + 4 \)
• C. \( 4x + \frac{1}{2x} = 8x^2 \)
• D. \( (2x^2 + 4) = (5 + x)(2x - 3) \)

Hint:

A quadratic equation is of the form \( ax^2 + bx + c = 0 \), where the highest power of \(x\) is 2.

Answer 1

The Correct Option is B

We need to identify which of the given equations is quadratic. A quadratic equation is a polynomial of degree 2, i.e., it involves \(x^2\) and no higher powers of \(x\). Option (A): \[ (x+1)(x-1) = x^2 - 4x^3. \] Expanding the left-hand side: \[ (x+1)(x-1) = x^2 - 1. \] Thus, the equation becomes: \[ x^2 - 1 = x^2 - 4x^3. \] Simplifying this: \[ x^2 - x^2 = -4x^3 + 1 \quad \Rightarrow \quad 0 = -4x^3 + 1. \] This is not a quadratic equation because it involves a cubic term (\(x^3\)). Option (B): \[ (x+4)^2 = 3x + 4. \] Expanding the left-hand side: \[ (x+4)^2 = x^2 + 8x + 16. \] Thus, the equation becomes: \[ x^2 + 8x + 16 = 3x + 4. \] Rearranging terms: \[ x^2 + 8x + 16 - 3x - 4 = 0 \quad \Rightarrow \quad x^2 + 5x + 12 = 0. \] This is a quadratic equation because it is of degree 2. Option (C): \[ 4x + \frac{1}{2x} = 8x^2. \] Multiplying through by \(2x\) to eliminate the fraction: \[ 8x^2 + 1 = 16x^3. \] This is not a quadratic equation because it involves a cubic term (\(x^3\)). Option (D): \[ (2x^2 + 4) = (5 + x)(2x - 3). \] Expanding the right-hand side: \[ (5 + x)(2x - 3) = 10x - 15 + 2x^2 - 3x = 2x^2 + 7x - 15. \] Thus, the equation becomes: \[ 2x^2 + 4 = 2x^2 + 7x - 15. \] Simplifying: \[ 2x^2 - 2x^2 + 4 = 7x - 15 \quad \Rightarrow \quad 4 = 7x - 15 \quad \Rightarrow \quad 7x = 19 \quad \Rightarrow \quad x = \frac{19}{7}. \] This is a linear equation, not a quadratic equation. Conclusion: The quadratic equation is option (B), \( (x+4)^2 = 3x + 4 \).