Question

$\huge\sf\fbox\purple{ \: \: Question \: \: }$

Zeroes of p(x) = x^2- 27 are:
a. ±9√3
b. ±3√3
c. ±7√3
d. None of the above

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Answer 1

Answer:

Given :-

• x² - 27.

To Find :-

• What are the zeroes of the polynomial.

Solution :-

Given Equation :

$\mapsto \sf\bold{\purple{x^2 - 27}}$

$\implies \sf P(x) =\: x^2 - 27$

So, let P(x) = 0 we get,

$\implies \bf x^2 - 27 =\: 0$

$\implies \sf x^2 - 27 =\: 0$

$\implies \sf x^2 =\: 27$

$\implies \sf x =\: ±\: \sqrt{27}$

$\implies \sf x =\: ±\: 3\sqrt{3}$

$\implies \sf\bold{\red{x =\: ±\: 3\sqrt{3}}}$

$\therefore$ The zeroes of the polynomial is ± 3√3.

Hence, the correct options is (b) ± 3√3 .

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EXTRA INFORMATION :-

$\clubsuit$ Sum Of Zeroes Formula :

$\longrightarrow \sf\boxed{\bold{\pink{Sum\: Of\: Zeroes\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}$

$\clubsuit$ Product Of Zeroes Formula :

$\longrightarrow \sf\boxed{\bold{\pink{Product\: Of\: Zeroes\: (\alpha\beta) =\: \dfrac{c}{a}}}}$

$\clubsuit$ Quadratic Polynomial Formula :

$\longrightarrow \sf\boxed{\bold{\pink{x^2 - (\alpha + \beta)x + \alpha\beta}}}$

Answer 2

Answer:

Solution is in the attachment.

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