Question:
Match the corresponding equations List-I with solutions List-II.
List-I List-II A. \(5x^2=0\) I. \(x=0,16\) B. \(2x^2-32=0\) II. \(x=\pm4\) C. \(2x^2+32=0\) III. \(x=0\) D. \(2x^2-32x=0\) IV. \(x=\pm4i\)
Match the corresponding equations List-I with solutions List-II.
| List-I | List-II |
|---|---|
| A. \(5x^2=0\) | I. \(x=0,16\) |
| B. \(2x^2-32=0\) | II. \(x=\pm4\) |
| C. \(2x^2+32=0\) | III. \(x=0\) |
| D. \(2x^2-32x=0\) | IV. \(x=\pm4i\) |
Show Hint
When \(x^2\) becomes negative, roots become imaginary. For example, \(x^2=-16\Rightarrow x=\pm4i\).
Updated On: May 22, 2026
- A-I, B-II, C-III, D-IV
- A-III, B-II, C-IV, D-I
- A-III, B-IV, C-II, D-I
- A-IV, B-III, C-II, D-I
Show Solution
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The Correct Option is B
Solution and Explanation
Concept: Quadratic equations are solved by factorisation or by isolating \(x^2\).
Step 1: Solve equation A.
\[ 5x^2=0 \] \[ x^2=0 \] \[ x=0 \] So, \[ A \rightarrow III \]
Step 2: Solve equation B.
\[ 2x^2-32=0 \] \[ 2x^2=32 \] \[ x^2=16 \] \[ x=\pm4 \] So, \[ B \rightarrow II \]
Step 3: Solve equation C.
\[ 2x^2+32=0 \] \[ 2x^2=-32 \] \[ x^2=-16 \] \[ x=\pm4i \] So, \[ C \rightarrow IV \]
Step 4: Solve equation D.
\[ 2x^2-32x=0 \] \[ 2x(x-16)=0 \] \[ x=0,\ 16 \] So, \[ D \rightarrow I \] Therefore, the correct matching is: \[ A-III,\ B-II,\ C-IV,\ D-I \]
Step 1: Solve equation A.
\[ 5x^2=0 \] \[ x^2=0 \] \[ x=0 \] So, \[ A \rightarrow III \]
Step 2: Solve equation B.
\[ 2x^2-32=0 \] \[ 2x^2=32 \] \[ x^2=16 \] \[ x=\pm4 \] So, \[ B \rightarrow II \]
Step 3: Solve equation C.
\[ 2x^2+32=0 \] \[ 2x^2=-32 \] \[ x^2=-16 \] \[ x=\pm4i \] So, \[ C \rightarrow IV \]
Step 4: Solve equation D.
\[ 2x^2-32x=0 \] \[ 2x(x-16)=0 \] \[ x=0,\ 16 \] So, \[ D \rightarrow I \] Therefore, the correct matching is: \[ A-III,\ B-II,\ C-IV,\ D-I \]
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