Question

Suppose the functions f(x)= -x2 +4x-p and g(x)=x2-px+9 has no real roots for some P belong to integer. Find the value of P.​

Answer 1

Answer:

The value of  $p$ is 5.

Step-by-step explanation:

Given the functions,

$f(x)=-x^2+4x-p$

and

$g(x)=x^2-px+9$

We need to find the value of $p$ so that these functions have no real roots.

For a quadratic equation of the form $ax^2+bx+c$ has no real roots if its discriminant $b^2-4ac$ is negative.

So we need to find the discriminant of the functions given.

$discriminant \ of \ f(x) =4^2-(4\times (-1)\times (-p))=16-4p\\\\discriminant \ of \ g(x) =(-p)^2-4\times 1\times 9=p^2-36$

For the function $f(x)$, the discriminant is negative implies,

$\implies 16-4p<0\\\\\implies 4p>16\\\\\implies p>\frac{16}{4}=4$

Similarly, For the function $g(x)$, the  discriminant is negative implies,

$\implies p^2-36<0\\\\\implies p^2<36\\\\\implies p<6 \ \ \ or\ \ \ p<-6$

So from these observations, $p$ should be greater than 4 and  less than 6. Since  $p$ is an integer, the value of  $p$ should be 5.