Let p, q \in R. If 2 – \sqrt{3} is a root of the quadratic equation, x2 + px + q = 0, then

Engineering

Mathematics

Quadratic Equation and Nature of Roots

Question

Let p, q $\in$ R. If 2 – $\sqrt{3}$ is a root of the quadratic equation, x² + px + q = 0, then

p² – 4q – 12 = 0

q² + 4p + 14 = 0

q² – 4p – 16 = 0

p² – 4q + 12 = 0

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In given question p,q $\in$ R. If we take other root as any real number a , then quadratic equation will be
$x^{2} - (α + 2 - \sqrt{3}) x + α . (2 - \sqrt{3}) = 0$

Now, we can have none or any of the options can be correct depending upon 'a'.

Instead of p,q $\in$ R it should be p,q $\in$ Q then other root will be $2 + \sqrt{3}$

$\Rightarrow p = - (2 + \sqrt{3} + 2 - \sqrt{3}) - - 4 and q = (2 + \sqrt{3}) (2 - \sqrt{3}) = 1$

⇒ p² – 4q –12 = (–4)² – 4 = 12

= 16 – 16 = 0

Option (1) is correct.