If the roots of the equations(given in the picture as question no. 3) are simultaneously real then prove that b2=ac


Hi, 1) ax²+2bx+c=0 have real roots then (2b)²-4ac>=0 ==> 4b²>=4ac ==> b²>=ac 2)  bx²-2√(ac)x+b=0 have real roots then  (-2√(ac))²-4b²>=0==> 4ac>=4b²==> ac>=b² So b²>=ac>=b²==> b²=ac

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