find the square root of 7+24i
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Let the square root of 7 + 24i = a +bi where a and b are real numbers.
We know that i^2 = -1
Now (a+bi)^2 = 7 + 24i
=> a^2 -b^2 + 2(a)(b)i =7 + 24i
Comparing both sides we get two equations a^2 - b^2 =7
And 2ab = 24
=> ab = 12
=> b =12/a
Now putting the value of b in equation 1 we get
a^2 - (12/a)^2 = 7
=> a^2 -(144/a^2) = 7
=> a^4 -144 = 7a^2
=> a^4 - 7 a^2 -144 = 0
Solving above quadratic equation we get
a^2 = (7+√(576+49))/2 or a^2 = (7-√(576+49))/2
=> a^2 = (7+25)/2 or a^2= (7–25)/2
=> a^2 = 16 or a^2 = -9
Since we have assumed a as real a^2 can't be negative
Therefore a^2 = 16
=> a = 4 or a = -4
Putting the value of a in ab = 12 we get
b = 3 or b = -3
Therefore our required solution is 4 + 3i or -4 -3i
Hope it helps!!!
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Answer:
Hence answer are :- +-(4+3i)
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