Question

Find the zeros of x square + 1/6 -2 and verify the relation between zeros and their coefficient​

Answer 1

Answer:

$\large{\underline{\boxed{\sf x = - \frac{ 3}{2} \: \: or \: \: x = \frac{4}{3} }}}$

Step-by-step explanation:

Given polynomial -

$\sf {x}^{2} + \frac{1}{6} x - 2 = 0 \\ \\ \implies \sf \frac{6x {}^{2} + x - 12 }{6} = 0 \\ \\ \implies \sf 6x {}^{2} + x - 12 = 0$

We can solve it, by splitting the middle term.

$\implies \sf 6x {}^{2} + x - 12 = 0 \\ \\ \implies \sf 6x {}^{2} + 9x - 8x - 12 = 0 \\ \\ \implies \sf 3x(2x + 3) - 4(2x + 3) = 0 \\ \\ \implies \sf (2x + 3)(3x - 4) = 0$

By zero product rule -

$\implies \sf (2x + 3 )= 0 \: or \: (3x - 4) = 0 \\ \\ \implies \bf x = - \frac{ 3}{2} \: \: or \: \: x = \frac{4}{3}$

Hence, the zeroes are $\bf x = - \frac{ 3}{2} \: or \: \frac{4}{3}$

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Verification of relationships:

$\bf Sum \: of \: zeros = \frac{ - (coefficient \: of \: x)}{coefficient \: of \: {x}^{2} } \\ \\ \implies - \frac{3}{2} + \frac{4}{3} = \frac{ - ( \frac{1}{6}) }{1} \\ \\ \implies \frac{ - 9 + 8}{6} = - \frac{1}{6} \\ \\ \implies - \frac{1}{6} = - \frac{1}{6} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \tt verified!}$

Now,

$\bf Product \: of\: Zeros = \frac{Constant\:term}{Coefficient\: of \:x^2}$

$\implies - \frac{3}{2} \times \frac{4}{3} = \frac{ - 2}{1} \\ \\ \implies \frac{ - 12}{6} = - 2 \\ \\ \implies - 2 = - 2 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \tt verified!}$