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

Step-by-step explanation:

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$\maltese \: \large \underline{ \bf{Answer:-}}$

Given :

$\bold {• \: \: \: S um \:of \: \: zeroes \: \: of \: \: a \: \: quadratic \: \: polynomial \: \: ( \alpha + \beta ) = 0 }$

$\bold {• \: \: \:product \: \:of \: \: zeroes \: \: of \: \: a \: \: quadratic \: \: polynomial \: \: ( \alpha \beta ) = - 1 }$

To Find :

• Zeroes of the polynomials

Solution :

General form for quadratic equations,

${\bf{ \boxed{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta }}}$

we have that ,

$\alpha + \beta = 0$

$\alpha \beta = - 1$

Substitute the values in the form ,

$\bf \implies \: {x}^{2} - (0)x -1$

$\implies \bf \: {x}^{2} - 1$

Next , find the zeroes of the polynomial.

It is in the form (a²-b²)

•°• (a²-b²) = (a+b)(a-b)

$\implies \sf \: {x}^{2} - {1}^{2}</u></strong><strong><u>$

$\implies \sf \:(x + 1)(x - 1) = 0$

$\implies \sf \:x = - 1 \: \: and \: \: 1$

•°• Zeroes of the polynomial is ±1 !!

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Answer 2

$\bold AnswEr:-$

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,

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