Question

Q.74. If one root of quadratic equation 2x2-px+q=0 is 2+√3 (p, q ∈ rational numbers), then pq is equal to
(1) 2
(2) 3
(3) 4
(4) 8

Answer 1

Given ,
one root of quadratic equation 2x² - px + q = 0 is 2 + √3
then, other root must be (2 - √3)

Now, sum of roots = -coefficient of x/coefficient of x²
(2 + √3) + (2 - √3) = -(-p)/2
⇒ 4 = p/2
⇒p = 8

product of roots = constant/Coefficient of x²
(2 + √3)(2 - √3) = q/2
⇒(2² - √3²) = q/2
⇒ 4 - 3 = q/2
⇒ 1 = q/2
⇒q = 2

∴ pq = 8 × 2 = 16 , your options are wrong . Answer should be 16

Answer 2

Hello Dear.

Here is your answer---


Given Conditions---

The Root of the Quadratic Equations 2x² - px + q = 0 is 2 + √3

∴ other roots of the given quadratic equations must be 2 -√3



From the given quadratic equation,

Coefficient of x² = 2

Coefficient of x = -p

Constant term = q

Thus,

For p.

Sum of roots of the quadratic equation = -Coffecient of x/Coeffecient of x²
                 ⇒     2 + √3 +2 - √3  = -(-p)/(2)
                 ⇒      4  = p/2
                  ⇒ p = 4 × 2
                  ⇒ p = 8.
For q.

Products of the given roots = Constant term/Coefficient of x²
                      (2 + √3)(2 - √3) = q/2
                         (2)² - (√3)² = q/2
                           4 - 3 = q/2
                               1 = q/2
                               q = 2

Thus, Products of p and q = p × q
                                          = 8 × 2
                                          = 16.

Thus, the answer is 16.
Options given in the questions are wrong.


Hope it helps.

Have a Marvelous Day.