Math, asked by santhiyaammu315, 1 day ago

maths pls answer pls step by step pls


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Answered by tennetiraj86
2

Given :-

The number 6-8i

To find :-

The square root of 6-8i

Solution :-

Given number = 6-8i

The square root of 6-8i = √(6-8i)

Let √(6-8i) = a + i b , Where a and b are real numbers

On squaring both sides then

=>[√(6-8i)]² = (a+ib)²

=> 6-8i = a²+2(a)(ib)+(ib)²

=> 6-8i = a²+i2ab+i²b²

=> 6-8i = a²+(2ab)i +(-b²)

Since, = -1

=> 6-8i = a²+(2ab)i -b²

=> 6-8i = (a²-b²) +(2ab)i

On comparing both sides then

a²-b² = 6 ----------(1)

and 2ab = -8

=> ab = -8/2

=> ab = -4

=> b = -4/a ---------(2)

On substituting the value of b in (1) then

=> a²-(-4/a)² = 6

=> a²-(16/a²) = 6

=> [(a²×a²)-16]/a² = 6

=> (a⁴-16)/a² = 6

=> a⁴-16 = 6a²

=> a⁴-6a²-16 = 0

=> a⁴-8a²+2a²-16 = 0

=> a²(a²-8)+2(a²-8) = 0

=> (a²-8)(a²+2) = 0

=> a²-8 = 0 or a²+2 = 0

=> a² = 8 or a² = -2

=> a² = 8 but a² ≠ 2

Since, a is a real number

=> a² = 8

=> a = ±√8

=> a = ±√(2×2×2)

=> a = ±2√2

Therefore, a = 2√2 and -2√2

If a = 2√2 then b = -4/(2√2)

=> b = -2/√2

=> b = (-√2×√2)/√2

=> b = -√2

If a = -2√2 then b = -4/(-2√2)

=> b = 2/√2

=> b = (√2×√2)/√2

=> b = √2

Therefore, a + ib = 2√2 -√2ior -2√2 +√2i

Therefore,

√(6-8i) = (2√2-√2 i) and -2√2 + √2 i

(or)

(2√2-√2 i) and -(2√2-√2 i)

Answer :-

The square root of 6-8i is (22-2 i) and -(22-2 i)

Answered by jaswantpanwar674
1

2√2+i√2. answer so this is the answer for you question

regards

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