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if P(x) = ax^2+bx+c and Q(x) = -ax^2 +dx+c , where ac not equal to 0 , then P(x)Q(x) = 0 has atleast two real roots. Is it true or false.plz explain also.

if P(x) = ax^2+bx+c and Q(x) = -ax^2 +dx+c , where ac not equal to 0 , then P(x)Q(x) = 0 has atleast two real roots. Is it true or false.plz explain also.

Grade:12

2 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
8 years ago
For the first equation without loss of generality assume a>0
then since none of a and c is zero

graph of P(x) will be -ve at -infinity and +ve at +inifinty
so it will cut x axis at once atleast

Same is the case with Q(x)

So totally the P(x)Q(x)=0 will have atleast 2 real roots

Arun Kumar
IIT Delhi
Askiitians Faculty
Anjali
13 Points
5 years ago
Discriminant of P(x) ;D`=b^2-4ac and that of Q(x) D"=d^2+4acWe are given that ac not equal to 0 .thus ac can either be -ve or + ve.If ac is -ve D` is +ve while D" can be -ve or + veAnd if ac is +ve D" is + ve and D` can be + ve or - ve.For both the cases P(x)Q(x)=0 has at least two real roots

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