find the zeros of the polynomial 6x²-x-2 and verify the relationship between the zeros and the Coefficients.​

1 thought on “find the zeros of the polynomial 6x²-x-2 and verify the relationship between the zeros and the Coefficients.​”

1. [tex]\huge\purple{\boxed{\underline{ANSWER}}}[/tex]

   GIVEN THAT:

   [tex]➲[/tex] Polynomial = 6x²-x-2

   TO FIND:

   [tex]➲[/tex] The zeros of given polynomial and also to verify relationship between the zeros and the Coefficients.

   FORMULA:

   [tex]➲[/tex] If any quadratic polynomial is in standard form ax²+ bx + c.

   where a and b are the Coefficients of x² and x respectively and c is the constant term

   if α and β are the zeros of this polynomial then

   [tex]&#10230 \: \: \alpha + \beta = \frac{ – b}{a} \\ \\ &#10230 \: \: \alpha \beta = \frac{c}{a} \: \: \: \: \: \: \: \: [/tex]

   SOLUTIONS:

   [tex]➲[/tex] polynomial = 6x²-x-2

   [tex]➲[/tex] After comparing its standard form ax²+ bx + c.

   [tex]➲[/tex] we get a = 6, b= -1, c = -2

   [tex]&#10230 \: \: 6 {x}^{2} – x – 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: 6 {x}^{2} – 4x + 3x – 2 \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: 2x(3x – 2) + 1(3x – 2) \\ \\ &#10230 \: \: (3x – 2)(2x + 1) \: \: \: \: \: \: \: \: \: \: \: \\ \\ if \: (3x – 2)(2x + 1) = 0 \\ \\ &#10230 \: \: 3x – 2 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: x = \frac{2}{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: 2x + 1 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: x = \frac{ – 1}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

   [tex]➲[/tex] now we get two zeroes of this polynomial

   [tex]&#10230 \: \: \alpha = \frac{2}{3} \: \: \: \: \: \\ \\ &#10230 \: \: \beta = \frac{ – 1}{2} [/tex]

   VERIFICATION:

   [tex]➲[/tex] Sum of the zeros

   [tex]&#10230 \: \: \alpha + \beta = \frac{ – b}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: \frac{2}{3} + (\frac{ – 1}{2}) = \frac{ – ( – 1)}{6} \\ \\ &#10230 \: \: \frac{2}{3} – \frac{1}{2} = \frac{1}{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \frac{4 – 3}{6} = \frac{1}{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: \frac{1}{6} = \frac{1}{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

   [tex]➲[/tex] Products of the zeros

   [tex]&#10230 \: \: \alpha \beta = \frac{c}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: \frac{\cancel2}{3} \times \frac{ – 1}{\cancel2} = \cancel\frac{ – 2}{6} \\ \\ &#10230 \: \: \frac{ – 1}{3} = \frac{ – 1}{3} \: \: \: \: \: \: \: \: [/tex]

   [tex]➲[/tex] HENCE VERIFIED.

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