Question

find the zeroes pf the polynomial x2-3 and verify the relationship between the zeroes and the coefficients..​

Answer 1

$\mathcal \blue{pic \: refers \: to \: attachment}$

Answer 2

Given

✭ We have the Polynomial x²-3

$\displaystyle\large\underline{\sf\blue{To \ Find}} </p><p>$

◈ The zeros of the equation?

◈ The relationship between zeros and coefficients?

$\displaystyle\large\underline{\sf\gray{Solution}} </p><p></p><p>$

So here we may use the identity,

$\displaystyle\sf \underline{\boxed{\sf x^2-y^2 = (x-y)(x+y)}} </p><p>$

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$\underline{\bigstar\:\textsf{According to the given Question:}} </p><p>$

In the given Equation we can write 3 as √3 × √3

$\begin{gathered}\displaystyle\sf\twoheadrightarrow x^2-3\\\end{gathered}$

$\begin{gathered}\displaystyle\sf\twoheadrightarrow x^2-\sqrt{3}^2\:\:\: \bigg\lgroup x^2-y^2 = (x-y)(x+y) \bigg\rgroup\\\end{gathered} </p><p>↠x </p><p>2</p><p> − </p><p>3</p><p> </p><p> </p><p>2$

$⎩</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎧</p><p> </p><p> x </p><p>2</p><p> −y </p><p>2</p><p> =(x−y)(x+y) </p><p>⎭$

$⎪</p><p>⎪</p><p>⎪</p><p>⎫</p><p> </p><p> </p><p>$

$\begin{gathered}\displaystyle\sf\twoheadrightarrow (x-\sqrt{3})(x+\sqrt{3}\\\end{gathered} </p><p>↠(x− </p><p>3</p><p> </p><p> )(x+ </p><p>3)$

$\displaystyle\sf\twoheadrightarrow\orange{x = \sqrt{3} \ or \ -\sqrt{3}}$

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So now the relationship of the zeros and the coefficients,

• a = 1
• b = 0
• c = -3

$</p><p>\begin{gathered}\displaystyle\sf\dashrightarrow \alpha+\beta = \dfrac{-b}{a}\\\end{gathered} </p><p>$

$</p><p>\begin{gathered}\displaystyle\sf\dashrightarrow \alpha+\beta = \dfrac{-0}{1}\\\end{gathered} </p><p> \\ ⇢α+β= </p><p>1</p><p>−0</p><p>$

$\displaystyle\sf\dashrightarrow \purple{\alpha+\beta = 0}$

Also,

$\begin{gathered}\displaystyle\sf\dashrightarrow \alpha\beta = \dfrac{c}{a}\\\end{gathered} </p><p>$

$\begin{gathered}\displaystyle\sf\dashrightarrow \alpha\beta = \dfrac{-3}{1}\\\end{gathered} </p><p></p><p>$

$\begin{gathered}\displaystyle\sf\dashrightarrow\pink{\alpha\beta = -3}\\\end{gathered}$

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