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find the zeroes of the quadratic polynomial 6x^2+x-2 and verify the relationship between the zeroes and the coefficients

User Ewbi
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Final answer:

To find the zeroes of the quadratic polynomial 6x²+x-2, we use the quadratic formula and determine the zeroes to be 1/2 and -3/2. We then verify the relationship between the zeroes and the coefficients of the polynomial, supporting it with Viète's formulas for the sum and product of roots.

Step-by-step explanation:

To find the zeroes of the quadratic polynomial 6x²+x-2, we can use the quadratic formula, which is derived from the equation ax²+bx+c = 0. The zeroes, or roots, of the quadratic are given by:

x = √{-b ± √{b²-4ac}}/{2a}

In this case, a = 6, b = 1, and c = -2. Plugging these into the quadratic formula, we get:

x = √{-1 ± √{(1)²-4(6)(-2)}}/{2(6)}

x = √{-1 ± √{1+48}}/{12}

x = √{-1 ± √{49}}/{12}

x = √{-1 ± 7}/{12}

So the two solutions are:

  • x = (√{-1 + 7})/{12} = 1/2
  • x = (√{-1 - 7})/{12} = -3/2

To verify the relationship between the zeroes and coefficients, we check the sum and product of the roots. According to Viète's formulas, the sum of the roots α and β is √{-b/a} and their product is c/a.

For the given quadratic:

Sum = α + β = 1/2 + (-3/2) = -1/6 = √{-b/a}

Product = α * β = (1/2) * (-3/2) = -3/4 = c/a

Here, √{-1}/{6} and -2/{6} (or -3/4) correspond with the coefficients of the quadratic polynomial, confirming the relationship between zeroes and coefficients.

User JustHooman
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