If one root of `x^2+px+1=0` is square that of the other thenp=

A

1, -2

B

3, -1

C

2, -5

D

2, 3

Verified by Experts

The correct Answer is:

A

Similar Questions

Explore conceptually related problems

If one root of x^(2) + px + 1 = 0 is the cube of the other root then p =

If one root of x^(2)+px+q=0 may be the square pf the other, then p^(3)+q^(2)+q =

If one root of x^(2)-x-k=0 is square of the other, then find the value of k.

If one root of the equation x^2+px+q=0 is the square of the other then

If one root of x^3 +3x^2 +5x +k=0 is sum of the other two roots then k=

If one root x^2-x-k=0 is square of the other, then k= a. 2+-sqrt(5) b. 2+-sqrt(3) c. 3+-sqrt(2) d. 5+-sqrt(2)

If one root of x^(2)+x +1 = 0 is (-1+sqrt(3)i)/(2) , then other root is :

Prove that the condition that one root of ax^(2)+bx+c=0 may be the square of the other is b^(3)+a^(2)c+ac^(2)=3abc .

If one root of the equation 8x^(2) - 6x-a-3=0 is the square of the other values of a are :

IF the sum of the square of the roots of x ^2+px -3=0 is 10 then the values of p=