JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 2)
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 – 16 = 0
q2 + 4p + 14 = 0
p2 – 4q + 12 = 0
Explanation
If a quadratic equation with rational coefficient has one irrational root then other root will be the conjugate of the irrational root.
Here x2 + px + q = 0 has one root 2 - $$\sqrt 3$$.
$$ \therefore $$ Other root will be 2 + $$\sqrt 3$$.
Sum of the roots = -p = 4
and product of the roots = q = 1
You can see p2 – 4q – 12 = 0 satisfy value of p = -4 and q = 1.
Here x2 + px + q = 0 has one root 2 - $$\sqrt 3$$.
$$ \therefore $$ Other root will be 2 + $$\sqrt 3$$.
Sum of the roots = -p = 4
and product of the roots = q = 1
You can see p2 – 4q – 12 = 0 satisfy value of p = -4 and q = 1.
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