IF p(x) =ax^2+bx+c and Q(x)=-ax^2+dx+c,ac is not equal to 0,then the equation p(x).q(x)=0 has

a.exactly 2 real roots b.atleast two real roots c.exactly 4 real roots d.no real roots

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

px=ax2+bx+cqx=-ax2+dx+cpx.qx=0ax2+bx+c-ax2+dx+c=0Consider,Ax2+Bx+C=0Discriminant=B2-4ACIn above equation if A and C are of opposite sign i.e. AC is negative. Then,B2-AC will be positive and hence equation will have real real roots.So if in any quadratic equation, coefficient of x2 and constant term are of opposite signAC<0, then thatequation will definitely have real roots.In px=ax2+bx+cProduct of coefficient of x2 and constant term=acIn qx=-ax2+dx+cProduct of coefficient of x2 and constant term=-acSo if ac is positive , then -ac will be negative and vice versa.One of the equation will definitely have real roots.ax2+bx+c-ax2+dx+c=0 will have atleast two real roots.Hence optionb is correct. 

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