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. 

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards

  • 13
What are you looking for?