Question

find the zeroes of the quadrature polynomial x^2 - 2√2 x and verify the relationship between the zeroes and the coefferents.​

Answer 1

$\large \underline{\pmb{\mathfrak{Question...}}}$

Find the zeroes of the quadratic polynomial $\sf x^2-2\sqrt{2}x$ and verify the relationship between the zeroes and coefficients.

$\large \underline{\pmb{\mathfrak{Solution...}}}$

Finding the factors of the polynomial :

$\: \: \: \sf \implies x ^{2} - 2 \sqrt{2} x = 0 \\ \\ \: \: \: \sf \implies x(x - 2 \sqrt{2} ) = 0 \\ \\ \: \: \: { \implies \pmb{ \sf{ x = 0 \: and \: 2 \sqrt{2} }}}$

Finding sum of the zeroes :

$\: \: \: \sf \implies Sum_{(zeroes)} = 0 + 2 \sqrt{2} \\ \\ \: \: \: \implies \pmb{\sf {Sum _{(zeroes)} = 2 \sqrt{2} } }$

Finding the product of the zeroes :

$\: \: \: \sf \implies Product_{(zeroes)} = 0(2 \sqrt{2} ) \\ \\ \: \: \: \implies\pmb{ \sf{ Product_{(zeroes)} = 0 }}$

When we compare $\sf x^2-2\sqrt{2}x$ to $\sf ax^2+bx+c,$ we get :

• $\sf a = 1$
• $\sf b = -2\sqrt{2}$
• $\sf c = 0$

Verifying the relationship between the zeroes and coefficients :

$\: \: \: \pmb{ \sf{ \alpha + \beta = \dfrac{ - b}{a}}} \\ \\ \: \: \: \sf \implies2 \sqrt{2} = \frac{- ( - 2 \sqrt{2} )}{1} \\ \\ \: \: \: \sf \implies2 \sqrt{2} = 2 \sqrt{2} \\ \\ \: \: \: {\pmb{\sf {Verified!}}}\\ \\ ~ \: \: \pmb{\sf{ \alpha \beta = \dfrac{c}{a}}} \\ \\ \: \: \: \sf \implies0 = \dfrac{0}{1} \\ \\ \: \: \: \sf \implies0 = 0 \\ \\ \: \: \: \pmb{\sf{ Verified!}}$