Question

find the quadratic poly. which the given no. asa the sum and product of zeros .1/4,-1

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Answer 1

$\huge\mathfrak\red{Answer:}$

Quadratic polynomial:

• Quadratic polynomial is a polynomial having highest power(degree) as 2.

Given:

• We have been given that the sum of zeroes is (-1/4) and product of zeroes is (-1).

To Find:

• We need to find the quadratic polynomial.

Solution:

As it is given that sum of zeroes(α + β) is (-1/4) and the product of zeroes(αβ) is (-1).

We can find the quadratic polynomial by this formula :

k[x² - (α + β)x + αβ]

Substituting the values, we have

k[x² - (-1/4)x + (-1)]

= k[x² + 1/4x - 1]

Hence the required quadratic polynomial is x² + 1/4x - 1.

Answer 2

Answer:

⠀⠀⋆ Sum of Zeroes $\rm(\alpha +\beta)={}^{1}\!/{}_{4}$

⠀⠀⋆ Product of Zeroes $\rm(\alpha\beta)=-\:1$

$\underline{\bigstar\:\textsf{Standard Form of Quadratic Polynomial :}}$

$\dashrightarrow\sf\:\:x^2-(Sum\:of\:Zeroes)x+(Product\:of\: Zeroes)=0\\\\\\\dashrightarrow\sf\:\:x^2-(\alpha +\beta)x+(\alpha\beta)=0\\\\\\\dashrightarrow\sf\:\:x^2 - \dfrac{1}{4}x + ( - 1) = 0\\\\\\\dashrightarrow\sf\:\:x^2 - \dfrac{x}{4} - 1 = 0\\\\\\\dashrightarrow\sf\:\: \dfrac{4x^2 - x - 4}{4} = 0\\\\\\\dashrightarrow\sf\:\:4x^2 - x - 4 = 0 \times 4\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf 4x^2 - x - 4 = 0}}$

⠀

$\therefore\:\underline{\textsf{Hence, required polynomial is \textbf{(4x ^\text2 - x - 4 = 0)}}}.$

$\rule{200}{2}$

⠀⠀⠀⠀⠀⠀⠀⠀Verification

$:\implies\sf 4x^2-x-4=0\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{By Discriminate Formula.}}\\\\:\implies\sf x=\dfrac{-\:b\pm\sqrt{b^2-4ac}}{2a}\\\\\\:\implies\sf x=\dfrac{-\:( - 1)\pm\sqrt{( - 1)^2-(4 \times 4 \times - 4)}}{2 \times 4}\\\\\\:\implies\sf x = \dfrac{1\pm\sqrt{1 + 64}}{8}\\\\\\:\implies\sf x = \dfrac{1 + \sqrt{65}}{8} \quad or \quad\dfrac{1 - \sqrt{65}}{8}$

$\rule{100}{0.9}$

★ Sum of Zeroes :

$\textsf{a = 4,\quad b = - 1,\quad c = - 4}\\\\\longrightarrow\sf (\alpha + \beta) = \dfrac{-b}{a} \\\\\\\longrightarrow\sf \dfrac{1 + \sqrt{65}}{8} +\dfrac{1 - \sqrt{65}}{8} = \dfrac{ - ( - 1)}{4}\\\\\\\longrightarrow\sf \dfrac{1 + \sqrt{65} + 1 - \sqrt{65} }{8} = \dfrac{1}{4}\\\\\\\longrightarrow\sf \dfrac{2}{8} = \dfrac{1}{4}\\\\\\\longrightarrow\underline{\boxed{\sf \dfrac{1}{4} = \dfrac{1}{4} }}$

⠀

★ Product of Zeroes :

$\longrightarrow\sf ( \alpha \beta) = \dfrac{c}{a}\\\\\\\longrightarrow\sf \dfrac{1 + \sqrt{65}}{8} \times \dfrac{1 - \sqrt{65}}{8} = \dfrac{ - 4}{4}\\\\\\\longrightarrow\sf \dfrac{(1)^2 - (\sqrt{65})^{2}}{64} = - 1\\\\\\\longrightarrow\sf \dfrac{1 - 65}{8} = - 1\\\\\\\longrightarrow\sf \dfrac{ - 64}{8} = - 1\\\\\\\longrightarrow\underline{\boxed{\sf -1 = - 1}}$