Question

Solution

Correct option is

5

Let α and β be two roots of x2 – px + q = 0 and assume α < β. we have α + β = p, αβ = q

As q is prime, we must have α = 1, β = q.

Thus, p = 1 + q ⇒ p – q = 1.

As p and q are prime, this is possible if p = 3, q = 2.

SIMILAR QUESTIONS

Q1

If x2 + px + 1 is a factor of ax2 + bx + c, then

Q2

If 8, 2 are the roots of x2 + ax + β = 0 and 3, 3 are the roots of x2 + α xb = 0 then the roots of x2 + ax + b = 0 are

Q3

The number of real roots of Q4

The product of real of the equation is

Q5

Sum of the non – real roots of  Q6

If tan A and tan B are the roots of the quadratic equation x2 – px + q = 0, then value of sin2 (A + B) is

Q7

If x Ïµ R, the number of solution of = 1 is

Q8

If lmn are real, l + m ≠ 0, then the roots of the equation

(l + m)x2 – 3(l – m)x – 2 (l + m) = 0 are

Q9

The real values of a for which the sum of the squares of the roots of the equation x2 – (a – 2) x – a – 1 = 0 assume the least value is

Q10

The real values of a for which the quadratic equation 3x2 + 2 (a2 + 1) x+(a2 – 3a + 2) = 0 possesses roots opposite signs lie in