Suppose a,b,c are three real numbers, such that the quadratic equation x²-(a+b+c)x+(ab+bc+ac)=0,has roots of the form α±iβ, where α>0 β≠0 are real numbers (i=sqrt(-1)). Show that (I) the numbers a,b,c are all positive. (II) the numbers √a,√b,√c form the sides of a triangle.
I tried to look at the roots of the quadratic equation by using the quadratic formula.
[(a+b+c)±√((a+b+c)²-4(ab+bc+ac))]/2 = (a+b+c)/2 ± √D/2
Here by comparison we have α=(a+b+c)/2 and β=√D/2 Now the discrimant has to be negative . Hence we have (a+b+c)²<4(ab+bc+ac)
(a²+b²+c²)<2(ab+bc+ca) Now a+b+c must me positive since It corresponds to α in α±iβ which is given to be positive..Now ab+bc+ac is also positive since it is greater than sums of the perfect squares of 3 numbers.so a+b+c ,ab+bc+ac postive and twice of ab+bc+ac is greater than a²+b²+c². This is all where I got.. I also understood what they mean by √a,√b,√c form triangle. It means that sum of any pair of square roots must exceed the the third square root. But I don't understand how it is