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+//answer match + roots
+
+clc;
+Pm=10000;
+V=400;
+f=50;
+smT=0.1;
+P=4;
+Ns=(120*f)/P;
+
+//for (i)
+disp('(i)');
+//As per given conditions the slip is given by equation Sfl2-0.4Sfl+0.01=0
+V=[1 -0.4 0.01];
+R=roots(V);
+Sfl=R(2);
+Nr=Ns*(1-Sfl);
+mprintf('The slip is %f \n The rotor speed is %f r.p.m',Sfl,ceil(Nr));
+
+//for (ii)
+disp('(ii)');
+Pg=Pm/(1-Sfl);
+Prot=Sfl*Pg;
+mprintf('The rotor ohmic loss is %f W \n',Prot);
+
+//for (iii)
+disp('(iii)');
+Tefl=Pg/(2*3.14*(Ns/60));
+Test=(4*Tefl)/((smT)+(1/smT));
+mprintf('starting torque is %f Nm \n',Test);
+
+//for (iv)
+disp('(iv)');
+a=sqrt(((Sfl*Sfl)+(smT*smT))/((Sfl)*(Sfl)*(1+(smT)*(smT))));
+mprintf('starting current = %f full load current\n',a);
+
+//for (v)
+disp('(v)');
+// answer is slightly different in book
+b=sqrt((1/2)*(1+(smT/Sfl)^2));
+mprintf('stator current at maximun torque = %f full load current \n',b);
+
+//for (vi)
+disp('(vi)');
+E=(Pm/Pg)*100;
+mprintf('full load efficiency is = %f percent\n',E);
+
+//for (vii)
+disp('(vii)');
+//As per given conditions
+smT1=3*smT;
+mprintf('New slip value is %f \n',smT1);
+
+//for (viii)
+disp('(viii)');
+//According to the given conditions s1(2)-1.2s+0.09
+VV=[1 -1.2 0.09];
+RR=roots(VV);
+s1=RR(2);
+Nr1=Ns*(1-s1);
+mprintf('full load slip is %f rotor speed is %f r.p.m',s1,Nr1);
+
+//for (ix)
+disp('(ix)');
+Test1=((2)/((1/0.3)+(0.3)))*(2*Tefl);
+mprintf('starting torque is %f Nm \n',Test1);
+
+//for (x)
+disp('(x)');
+c=sqrt((s1^2+smT1^2)/(s1^2*(1+smT1^2)));
+mprintf('starting current = %f full load current \n',c);
+
+//for (xi)
+disp('(xi)');
+Protfl=s1*Pg;
+mprintf('Rotor ohmic loss at full load torque is %f W \n',Protfl);
+
+//for (xii)
+disp('(xii)');
+Pm1=(1-s1)*Pg;
+E=Pm1/Pg;
+mprintf('Efficiency is %f percent',E*100);