Question

obtain the Zeroes of the polynomial root 3x^2 - 8 x + 4 root 3 and verify the relation between it's zeroes and coefficient. no spam spammers be away correct✅ answer needed no copies​

Answer 1

let f(x)= 3 x 2 −8x+4 3 by splitting the middle term, we get f(x)= 3 x 2 −6x−2x+4 3 = 3 x(x−2 3 )−2(x−2 3 ) =( 3 x−2)(x−2 3 ) on putting f(x)=0, we get ( 3 x−2)(x−2 3 )=0 ⇒ 3 x−2=0 or x−2 ( 3=0) ⇒x= 3 2 orx=2 3 thus, the zeroes of the given polynomial 3 x 2 −8x+4 3 are 3 2 and 2 3 verification : sum of zeroes =α+β= 3 2 +2 3 = 3 2+6 = 3 8 or =− coefficient of x 2 coefficient of x =− 3 (−8) = 3 8 product of zeroes =αβ= 3 2 ×2 3 =4 or = coefficient of x 2 constant term = 3 4 3 =4 so, the relationship between the zeroes and the coefficients is verified.

Answer 2

Answer:

→ Hey Mate,

→ Given Question : obtain the Zeroes of the polynomial root 3x^2 - 8 x + 4 root 3 and verify the relation between it's zeroes and coefficient.

→ Step-by-step explanation:

→ Answer: -  $\sqrt{3} x^2 - 8 x + 4\sqrt{3} = 0$               ∵   $( a x^2 + bx + c= 0)$

→                   $\sqrt{3}x^2 - 6x -2x + 4\sqrt{3} = 0$

→                   $\sqrt{3} x ( x - 2\sqrt{3} ) -2 ( x - 2\sqrt{3} )=0$

→                   $( \sqrt{3 } x-2 ) ( x -2 \sqrt{3} ) = 0$

→                   $x = \frac{2}{\sqrt{3} }$      $x = 2\sqrt{3}$

→                 $a + B = \frac{-b}{a} = -(\frac{-8}{\sqrt{3} } )= \frac{8}{\sqrt{3} }$

→                $a B= \frac{c}{a } = 4\frac{\sqrt{3} }{\sqrt{3} } = 4$  Here $\sqrt{3}$  $and$   $\sqrt{3}$ gets cancelled.

→                α + β

→             $\frac{2}{\sqrt{3} } + 2\sqrt{3}$

→            $\frac{2+6}{\sqrt{3} } = \frac{8}{\sqrt{3} }$

→          αβ= $\frac{2}{\sqrt{3} }$   ×  $2\sqrt{3}$      [ ∵ Here $\sqrt{3}$  and $\sqrt{3}$ gets cancel ]

→        4 is the solution.

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