Obtain the zeros of p(x) = 6x^2 - x - 2. Also | vedclass

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(B) To find the zeros of $p(x) = 6x^2 - x - 2$,we set $p(x) = 0$.$6x^2 - x - 2 = 0$$6x^2 - 4x + 3x - 2 = 0$$2x(3x - 2) + 1(3x - 2) = 0$$(2x + 1)(3x -

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$2/3, -1/2$

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(B) To find the zeros of $p(x) = 6x^2 - x - 2$,we set $p(x) = 0$.$6x^2 - x - 2 = 0$$6x^2 - 4x + 3x - 2 = 0$$2x(3x - 2) + 1(3x - 2) = 0$$(2x + 1)(3x - 2) = 0$Thus,the zeros are $x = -1/2$ and $x = 2/3$.Let $\alpha = 2/3$ and $\beta = -1/2$.For a quadratic polynomial $ax^2 + bx + c$,the sum of zeros is $-b/a$ and the product of zeros is $c/a$.Here,$a = 6, b = -1, c = -2$.Sum of zeros: $\alpha + \beta = 2/3 + (-1/2) = (4 - 3)/6 = 1/6$. Also,$-b/a = -(-1)/6 = 1/6$. Thus,$\alpha + \beta = -b/a$.Product of zeros: $\alpha \cdot \beta = (2/3) \cdot (-1/2) = -2/6 = -1/3$. Also,$c/a = -2/6 = -1/3$. Thus,$\alpha \cdot \beta = c/a$.Hence,the relationship is verified.

Obtain the zeros of $p(x) = 6x^2 - x - 2$. Also,verify the relationship between the zeros and the coefficients of $p(x)$.

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