3sqrt(- 1 )
= 3i
--------
//not sure if it is correct bool isomorphic(struct Node* root1,struct Node* root2) { if(root1 root2->value) return ( isomorphic(root1->left,root2->left) && isomorphic(root1->right,root2->right) isomorphic(root1->right,root2->left) && isomorphic(root1->left,root2->right) ); else return false; }
#include <stdio.h> #include <math.h> /* bear in mind that this has no idiot-proofing whatsoever. For example, it doesn't check to see if the numbers entered are positive (and sqrt doesn't handle imaginary numbers) */ int main(int argc, char *argv[]){ float val1, val2, root1, root2; printf("Gimme a number: "); scanf("%f", &val1); printf("Gimme another number: "); scanf("%f", &val2); root1 = sqrt(val1); root2 = sqrt(val2); printf("The square root of %f is %f.\n", val1, root1); printf("The square root of %f is %f.\n", val2, root2); printf("The product of %f and %f is %f\n", root1, root2, root1 * root2); return 0; }
implicit double precision(a-h,o-z) write(*,*) "please provide the a,b,c coeff" read(*,*) A,B,C D=B*B-4*A*C if(D.GT.0) then root1=(-B/(2*A))+(SQRT(D))/(2*A) root2=(-B/(2*A))-(SQRT(D))/(2*A) write(*,*) root1,root2 elseif(D.EQ.0) then root1=(-B/(2*A)) root2=root1 write(*,*) root1,root2 else root1=(-B/(2*A))+(SQRT(-D))/(2*A) root2=(-B/(2*A))-(SQRT(-D))/(2*A) a=(root1+root2)/2 b=(root1-root2)/2 write(*,*) 'realpartroot=',a, 'complexpartroot=',b endif stop END
square root 2 times square root 3 times square root 8
729 does not have "a square root of 3 times 3 times 3 times 327" so the question does not arise. 729 is a square number with a square root of 3 x 3 x 3 because (3 x 3 x 3) x (3 x 3 x 3) = 729
3.46410162
The square of 3 is 2.25 times the square of 2
It works out as -26 times the square root of 3
It is the same as: 5 times square of 3
3(3 square root of 2) = 9(square root of 2)
7.4915627415691541308549385962873
The square root of 9 is 3 or -3.3 times 3 is 9.-3 times -3 is also 9.So they are both the square root of 9.