How do you multiply $(n^2)/(2)-(5n)/(6) -2 =0$ ?

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Answer 1 · 2015-05-09T12:15:06

$n^2/2$ $- (5n)/6$ $- 2 = 0$

Here the L.C.M = 6

$(n^2 xx 3)/(2 xx 3)$ $- (5n)/6$ $- (2xx6)/(1 xx6) = 0$

$(3n^2)/(6)$ $- (5n)/6$ $- (12)/(6) = 0$
$(3n^2 - 5n - 12)/(6) = 0$

$3n^2 - 5n - 12 =6 xx 0$

$3n^2 - 5n - 12 = 0$

factoring by grouping / splitting the middle term:
$3n^2 - 5n - 12 = 0$

( $3 xx -12 = -36$ , $-36 = -9 xx 4$ and $-5$ = $4-9$ )

so,
$3n^2 - 9n + 4n - 12 = 0$

taking out the common terms:
$3n( n -3) + 4(n -3) = 0$
on grouping:
$(3n+4) (n - 3) = 0$

here, $n$ has two solutions :
$n = 3 and n = -4/3$

How do you multiply (n^2)/(2)-(5n)/(6) -2 =0(n^2)/(2)-(5n)/(6) -2 =0?