Question

Sum of the non – real roots of 

 

Solution

Correct option is

– 1

Put x2 + x = y [See 2.16 in Theory], so that the equation (1) becomes

     (y – 2) (y – 3) = 12

⇒  y2 – 5y – 6 = 0

⇒ (y – 6) (y + 1) = 0 ⇒ y = 6, – 1.

When y = 6 we get x2 + x – 6 = 0

⇒ (x + 3) (x – 2) = 0 or  x = – 3, 2

When y = – 1, we get x2 + x + 1 = 0

⇒       x = ww2 and their sum is – 1.

SIMILAR QUESTIONS

Q1

A quadratic equation with rational coefficients can have

Q2

If the roots of ax2 + bx + c = 0 are in the ratio mn then

Q3

For real x, the expression [(x + m)2 – 4mn]/[2(x – n)] can be have any value except

Q4

If the expression y2 + 2xy + 2x + my – 3 can be resolved into two rational factors, then m must be

Q5

If one root of the equation x2 + px + 12 = 0 is 4, while the equation x2 +px + q = 0 has equal roots, the value of q is

Q6

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

Q7

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

Q8

The number of real roots of 

Q9

The product of real of the equation  

              is

Q10

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