Math, asked by basur, 10 months ago

The zeros of the polynomial x^2-1 is

Answers

Answered by sabrinanandini2

$\huge{\textbf{ANSWER-}}$

x² - 1 = 0

x² = 1

x = √1

⇒ x = ± 1

The zeroes of x²-1 are 1 or -1

$\mathbb{ALTERNATIVE}$

The polynomial x²-1 is of the form a² - b²

As we know,

$\boxed{a^{2}-b^{2}=(a+b)(a-b)}$

Using this, we get

x² - 1 = 0

(x+1)(x-1) = 0

⇒ x+1 = 0 ⇒ x = -1

⇒ x-1 = 0 ⇒ x = 1

Hence, the zeroes of x²-1 are -1 and 1

$\texttt{The\:zeroes\:are\:1\:and\:-1}$

ck233: hi

sabrinanandini2: Heya

Answered by abdul9838

$<b> <body bgcolor ="skyblue">$

$\small \bf \green{hey \: mate \: here \: is \: ur \: ans} \\ \\ \small \bf \green{ \huge \: solution} \\ \\ \small \bf \green{given \: that} \\ \\ \small \bf \green{ {x}^{2} - 1 } \\ \\ \small \bf \green{as \: we \: can \: write} \\ \\ \small \bf \green{( {x})^{2} - (1)^{2} } \\ \\ \small \bf \green{ \underline{using \: this \: identity}} \\ \\ \small \bf \green{ {a}^{2} - {b}^{2} = (a + b)(a - b)} \\ \\ \small \bf \green{((x + 1)(x - 1)} \\ \\ \small \bf \green{x = - 1} \\ \\ \small \bf \green{x = 1} \\ \\ \small \bf \green{ \underline{zeros \: are \: - 1 \: and \: 1}}$