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what is number of quadratic equation that are unchanged by squaring their roots is There are four such cases x2 =0 root 0 (x-1)2=0 root 1 x(x+1)=0 roots 0 and 1 x2+x+1=0 roots ω and ω2 let x2 +bx +c=0 ............1 not let another equation whoose roots are square of this equation so X=x2 or x=√X put value of x X +b√X +c =0 or X +c =-b√X square X2 +c2 +2cX =b2X X2 +(2c-b2)X + c2 =0..................2 root of equation 2 is the square of root of equation 1 both equation will be same if their coefficient are in proportion 1/1 =b/(2c-b2) =c/c 2 b= 2c-b2 ................3 c=c2 ....................3 from equation 3 c=0 or 1 for c =0 b= 0 and -1 for c=1 b= 1 and -2 so four combination are possible
what is number of quadratic equation that are unchanged by squaring their roots is
There are four such cases x2 =0 root 0
(x-1)2=0 root 1
x(x+1)=0 roots 0 and 1
x2+x+1=0 roots ω and ω2
let x2 +bx +c=0 ............1
not let another equation whoose roots are square of this equation
so X=x2 or x=√X
put value of x
X +b√X +c =0
or X +c =-b√X
square
X2 +c2 +2cX =b2X
X2 +(2c-b2)X + c2 =0..................2
root of equation 2 is the square of root of equation 1
both equation will be same if their coefficient are in proportion
1/1 =b/(2c-b2) =c/c 2
b= 2c-b2 ................3
c=c2 ....................3
from equation 3 c=0 or 1
for c =0 b= 0 and -1
for c=1 b= 1 and -2
so four combination are possible
1. Consider the following differential equation with initial conditions: t 2 x'' + 5 t x' + 3 x = 0, x(1) = 3, x'(1) = -13. Assume there is a solution of the form: x (t) = t
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