Math, asked by ushamourya99, 3 months ago

B. Determine whether the pairs of numbers given below are co-prime numbers 50..55​

Answers

Answered by Anonymous
2

Answer:

Solve for p:

p^2 (-p^2 + h^2) = a^2 h^2

Expand out terms of the left hand side:

-p^4 + h^2 p^2 = a^2 h^2

Subtract a^2 h^2 from both sides:

-a^2 h^2 + h^2 p^2 - p^4 = 0

Substitute x = p^2:

-a^2 h^2 + h^2 x - x^2 = 0

Multiply both sides by -1:

a^2 h^2 - h^2 x + x^2 = 0

Subtract a^2 h^2 from both sides:

x^2 - h^2 x = -a^2 h^2

Add h^4/4 to both sides:

x^2 - h^2 x + h^4/4 = h^4/4 - a^2 h^2

Write the both sides as a square:

(x - h^2/2)^2 = -1/4 h^2 (-h + 2 a) (h + 2 a)

Take the square root of both sides:

x - h^2/2 = 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a)) or x - h^2/2 = -1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))

Add h^2/2 to both sides:

x = h^2/2 + 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a)) or x - h^2/2 = -1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))

Substitute back for x = p^2:

p^2 = h^2/2 + 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a)) or x - h^2/2 = -1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))

Take the square root of both sides:

p = sqrt(h^2/2 + 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))) or p = -sqrt(h^2/2 + 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))) or x - h^2/2 = -1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))

h^2 (-h + 2 a) (h + 2 a) = -h^4 + 4 a^2 h^2:

p = sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = -sqrt(h^2/2 + 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))) or x - h^2/2 = -1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))

h^2 (-h + 2 a) (h + 2 a) = -h^4 + 4 a^2 h^2:

p = sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = -sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or x - h^2/2 = -1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))

Add h^2/2 to both sides:

p = sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = -sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or x = h^2/2 - 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))

Substitute back for x = p^2:

p = sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = -sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p^2 = h^2/2 - 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))

Take the square root of both sides:

p = sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = -sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = sqrt(h^2/2 - 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a))) or p = -sqrt(h^2/2 - 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a)))

h^2 (-h + 2 a) (h + 2 a) = -h^4 + 4 a^2 h^2:

p = sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = -sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = sqrt(h^2/2 - 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = -sqrt(h^2/2 - 1/2 i sqrt(h^2 (-h + 2 a) (h + 2 a)))

h^2 (-h + 2 a) (h + 2 a) = -h^4 + 4 a^2 h^2:

Answer: |

| p = sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = -sqrt(h^2/2 + 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = sqrt(h^2/2 - 1/2 i sqrt(-h^4 + 4 a^2 h^2)) or p = -sqrt(h^2/2 - 1/2 i sqrt(-h^4 + 4 a^2 h^2))

Answered by Sriramgangster
5

 \: \: \: \: \: \: \:

Similar questions