Question

find the quadratic polynomial the sum of whose zeros is 0 and their product is -1 hence, find the zeros of the polynomial​

Answer 1

AnswEr:-

Quadratic polynomial = x² - 1

Zeroes of polynomial = -1 & 1

$\rule{200}{1}$

Explanation:-

Let the zeroes of polynomial be α & β

Given:-

• α + β = 0
• αβ = -1

We know the standard form of a quadratic polynomial:-

$\star\: \boxed{\boxed{\sf\blue{x^2 - (Sum \: of \: zeros)x + Product\: of\: zeros }}}$

Here,

⇒ Polynomial = x² - (0)x + (-1)

⇒ Polynomial = x² - 0x - 1

⇒ Polynomial = x² - 1

Therefore,

$\therefore\underline{\textsf{Quadratic polynomial = {\textbf{x ^2 - 1}}}}$

$\rule{200}{1}$

We got the quadratic polynomial as x² - 1

Now we can find the zeros of polynomial by factorization method:-

⇒ x² - 1 = 0

[We know, a² - b² = (a + b)(a - b) ]

⇒ (x + 1)(x - 1) = 0

⇒ x = -1 or x = 1

Therefore,

$\therefore\underline{\textsf{Zeros of polynomial = {\textbf{-1 \& 1 }}}}$

Answer 2

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Question :

Find the quadratic polynomial the sum of whose zeros is 0 and their product is -1 hence, find the zeros of the polynomial.

Solution :

$\tt{\red{Given \: - }}$

• Sum of Zeroes = 0

• Product of Zeroes = -1

A Quadratic Polynomial can be expressed as :

$\tt{ \purple{ \implies{ {x}^{2} - (sum \: of \: zeroes) + (product \: of \: zeroes) }}}$

Hence The required Polynomial becomes :

$\tt{ \orange{ \implies{ {x}^{2} - 0x - 1 = > {x}^{2} - 1}}}$

Factoring The Above Expression :

$\tt{\green{\implies{ (x-1)(x+1) }}}$

The Required Zeroes are 1 and -1.