यदि समीकरण की जड़ों में से एक |(7,6,x^(2)-25),(2,x^(2)-25,2),(x^(2)-25,3,7) |=0 x=3 है,...

If one of the roots of the equation |(7,6,x^(2)-25),(2,x^(2)-25,2),(x^(2)-25,3,7)|=0 is x=3 , then the sum of all other five roots is

7(x-2y)^(2)-25(x-2y)+12

int(x^(2)+3)/(sqrt(x^(2)-25))dx

7(x-2y)^2-25(x-2y)+12

(x+3)/(7)-(2x-5)/(3)=(3x-5)/(5)-25

(x-5)(x-6)=(25)/((24)^(2))

x ^ (2) + y ^ (2) = 25, x ^ (3) + y ^ (3) = 91

(x+5)/(2)+(x-5)/(3)=(25)/(6)

Solve :(3x)/(10)+(2x)/(5)=(7x)/(25)+(29)/(25)

Solve: (3x)/(10)+(2x)/5=(7x)/(25)+(29)/(25)