Question

solve the following equation by cardan method ..x³+ 3x-14 =0​

Answer 1

Answer:

f(x)=x3+3x−14  compulsorily has a real integral solution upon three solutions.

All good ?

We're under some conditions.  x>0∵f(−n)<0  where  n>0.  

f(x)  can be rewritten as ,  x(x2+3)=14  

14  is  1×14,2×7  (we're excluding negative combinations).

So ,  x(x2+3)=1(14)  or  x(x2+3)=2(7)  

But  x(x2+3)=1(14)  is not a solution because when  x=1 , x2+3≠14.  

∴x=2  is our integral solution.

So ,  (x−2)  is a factor of  f(x).  

Now,  f(x)=(x−2)(x2+2x+7)  

Using the quadratic equation,  x=−b±b2−4ac−−−−−−−√2a  , two other solutions of  f(x)  can be conveniently found out.

x=−1±6–√i  

Step-by-step explanation: