Question

2. Determine if 3 is a zero of the polynomial
$p(x) = \sqrt{x {}^{2} - 4x + 3 } + \sqrt{x {}^{2} - 9} \\ - \sqrt{4x {}^{2} - 14x + 6}$


Standard:- 10

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Answer 1

$Given p(x) : \sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} - \sqrt{4x^2- 14x + 6}$

We know that x = 3 is zero of polynomial, p(x), if p(3) = 0.

$= \ \textgreater \ \sqrt{(3)^2 - 4(3) + 3} + \sqrt{(3)^2 - 9} - \sqrt{4(3)^2 - 14(3) + 6}$

$= \ \textgreater \ \sqrt{9 - 12 + 3} + \sqrt{9 - 9} - \sqrt{36 - 42 + 6}$

$= \ \textgreater \ \sqrt{0} + \sqrt{0} + \sqrt{0}$

= > 0.



Therefore 3 is the zero of the polynomial!.