If a,b,c,d​ ∈ R, then show that the equation (x2+ax-3b) *(x2-cx+b) *(x2-dx+2b)=0 has atleast two real roots.

Dear Student ,
Please find below the solution to the asked query :

x2+ax-3b x2-cx+b x2-dx+2b=0Now,for x2+ax-3b=0D=a2-41-3b=a2+12bFor x2-cx+b=0D=-c2-41b=c2-4bFor x2-dx+2b=0D=-d2+412b=d2+8bLet us assume that x2+ax-3b has imaginary roots.Then,a2+12b<012b<-a2b<-a212b is negativeSince b is negative then -4b will be positive and c2 is positive as square of real numberis always positive.So, c2-4b will be positivex2-cx+b=0 has real rootsx2+ax-3b x2-cx+b x2-dx+2b=0 has atleast two real roots .
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