find the zeros of the quadratic equation f(x)= 6x²-3, and verify the relation between its zeros and coefficients. using the formula

Question

find the zeros of the quadratic equation f(x)= 6x²-3, and verify the relation between its zeros and coefficients.
using the formula
alpha+beta= -b/a
alpha × beta = c/a​

in progress 0
Nick 3 years 2021-08-06T00:04:33+00:00 2 Answers 16 views 0

Answers ( )

  1. sori
    0
    2021-08-06T00:06:25+00:00
    Reply

    Answer:

    Step-by-step explanation:

    We are already given the following formulas :-

    \blue{\alpha + \beta = \dfrac{-b}{a}}

    \red{\alpha \times \beta = \dfrac{c}{a}}

    Given that the polynomial is :-

    \sf{6x^2 + 0x - 3}

    Now, we will use factorisation as our first weapon :-

    6x² -3 = 0 [As we have to find the zeroes of the equation]

    6x² = 3

    x² = \dfrac{3}{6}

    x² = \dfrac{1}{2}

    x = \dfrac{1}{\sqrt{2}}

    Now, we can take one zero as 1/Root 2 and other -1/root 2

    \dfrac{1}{\sqrt{2}} + \dfrac{-1}{\sqrt{2}} | \dfrac{0}{6} = 0

    0 |0 Hence verified.

    \dfrac{1}{\sqrt{2}} \times \dfrac{-1}{\sqrt{2}} | \dfrac{-3}{6} = 0

    \dfrac{1}{2} | \dfrac{1}{2}

    Hence, proved.

  2. bonexptip
    0
    2021-08-06T00:06:31+00:00
    Reply

    Answer:

    f(x) = 6x²-3

    f(x) = 0

    6x²-3 = 0

    6x² = 3

    x² = 3/6

    x² = 1/2

    x = 1/√2 , -1/√2

    Roots of f(x) = 1/√2 , -1/√2

    Verification = (i) 6(1/√2)² – 3

    = 6(1/2) – 3

    = 3-3

    = 0

    (ii) 6(-1/√2)² – 3

    = 6(1/2) – 3

    = 3-3

    = 0

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