Match List-I with List-II.
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(x^{2}-4=0\) | (I) | \(x=\frac{7}{2}\) |
| (B) | \(2x-7=0\) | (II) | Quadratic equation |
| (C) | \(\frac{1}{x^{2}}-1=0\) | (III) | \(x=2\) and \(-2\) |
| (D) | \(4x^{2}+4x+1=0\) | (IV) | \(x=\pm1\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-I, C-IV, D-II
Each equation is solved for \(x\); an equation of degree 2 whose roots are not listed is described by its type.
(A) \(x^{2}-4=0\) — this matches (III) \(x=2\) and \(-2\). \(x^2=4\), so \(x=2\) or \(x=-2\).
(B) \(2x-7=0\) — this matches (I) \(x=\frac{7}{2}\). \(2x=7\), so \(x=\frac{7}{2}\).
(C) \(\frac{1}{x^{2}}-1=0\) — this matches (IV) \(x=\pm1\). \(\frac{1}{x^2}=1\), so \(x^2=1\) and \(x=\pm1\).
(D) \(4x^{2}+4x+1=0\) — this matches (II) Quadratic equation. Its highest power of \(x\) is 2, so it is a quadratic equation; it is \((2x+1)^2=0\), with the repeated root \(x=-\frac{1}{2}\), which is not among the roots in List-II.