If $(2 + i \sqrt{3})$ is a root of the equation $x^{2} + p x + q = 0$ , where p and q are real, then (p, q) equals to

(4, 7)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- 4, - 7)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- 4, 7)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(4, - 7)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $(- 4, 7)$
Since $2 + i \sqrt{3}$ is one root, then other root will be $2 - i \sqrt{3}$ .
$∴ x^{2} + p x + q = 0$ is given equatiion
$∴$ Sum of roots $= 2 + i \sqrt{3} + 2 - i \sqrt{3} = p$
$∴ p = - 4$
Product of roots $q = 4 + 3 = 7$

Suggest Corrections

Similar questions

If $2 + i \sqrt{3}$ is a root of the equation $x^{2} + p x + q = 0$ , where p and q are real, then (p, q) =

2 + i \sqrt{3}

x^{2} + p x + q = 0

If $(2 + i \sqrt{3})$ is a root of the equation $x^{2} + p x + q = 0$ where $p$ and $q$ are real, then find $(p, q)$ .

i \sqrt{3}

x^{2} + p x + q = 0

α

β

x^{2} + p x + q = 0

3 α + 4 β = 7

5 α - β = 4

(p, q)

Join BYJU'S Learning Program