Question

find the roots of given quatratic equation by using quatratic 2x2+x-4=40​

Answer 1

Step-by-step explanation:

ve seen three answers, but all of them give only the two real solutions. The equation given is equivalent to a polynomial equation of degree six, which must have six complex (not necessarily different) roots.

x3+1x3=18

⇒

(x3)2–18(x3)+1=0 .

Solving the (quadratic) equation for x3 , we have

x3=9±81–1−−−−√=9±45–√ .

So far, I’ve done the same work as my colleagues did.

Now, the cubic root. Defined as a real function, x=y√3 is the unique real number which satisfies x3=y . But, if y≠0 , this equation has three complex solutions

x1x2x3=y√3=y√3(−12+i3–√2)=y√3(−12−i3–√2),

which are three vectors in the complex plane making 120º angles between them.

Therefore, the six solutions of the equation given are

x1x2x3x4x5x6=9+45–√−−−−−−−√3=9+45–√−−−−−−−√3(−12+i3–√2)=9+45–√−−−−−−−√3(−12−i3–√2)=9−45–√−−−−−−−√3=9−45–√−−−−−−−√3(−12+i3–√2)=9−45–√−−−−−−−√3(−12−i3–√2).