JEE Main 2020 · 2 September · evening · Single correct · +4 / −1
The imaginary part of
( 3 + 2√(- 54) )^(1/2) - ( 3 - 2√(- 54) )^(1/2) can be :
- (a) -2√6
- (b) 6
- (c) √6
- (d) -√6
Answer and solution
Answer (a) -2√6
Solution
Maths sir · 3 steps
Write each number as a perfect square (a+bi)², and remember every square root has two signs.
- Clean up the root. √-54 = √54 i = 3√6 i, so 2√-54 = 6√6 i. The two numbers are 3 + 6√6 i and 3 - 6√6 i.
- Find the square root. Let (a+bi)² = 3 + 6√6 i. Then a² - b² = 3 and 2ab = 6√6, so ab = 3√6. Trying a = 3 gives b = √6, and 9 - 6 = 3 checks out.
√(3 + 6√6 i) = ±(3 + √6 i), √(3 - 6√6 i) = ±(3 - √6 i)
First root Second root Difference Imaginary part 3 + √6 i 3 - √6 i 2√6 i 2√6 3 + √6 i -3 + √6 i 6 0 -3 - √6 i 3 - √6 i -6 0 -3 - √6 i -3 + √6 i -2√6 i -2√6 - So. The imaginary part can be 2√6, 0 or -2√6. Of the options, only (a) -2√6 is possible.
The idea to keep. To find √(x+iy), set it equal to a+bi and match real and imaginary parts. Keep both signs of each root, because "can be" means checking every combination.
Answer from the NTA final answer key. Solution by PrepWiser.