How do you solve $21 n^{2} + 5 n - 6 = 0$ $21n ^ { 2} + 5n - 6= 0$ ?

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How do you solve $21 n^{2} + 5 n - 6 = 0$ $21n ^ { 2} + 5n - 6= 0$ ?

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Answer 1 · 2018-03-28T12:49:17

$n = \frac{3}{7}, - \frac{2}{3}$ $n=3/7, -2/3$

Explanation:

Given:
$\times x 21 n^{2} + 5 n - 6 = 0$ $color(white)(xxx)21n^2+5n-6=0$

$\Rightarrow 21 n^{2} - 9 n + 14 n - 6 = 0$ $rArr 21n^2-9n+14n-6=0$ [Broke up $5 n$ $5n$ as $- 9 n + 14 n$ $-9n +14n$ ]

$\Rightarrow 3 n (7 n - 3) + 2 (7 n - 3) = 0$ $rArr 3n(7n-3)+2(7n-3)=0$ [Grouping the Like Terms]

$\Rightarrow (7 n - 3) (3 n + 2) = 0$ $rArr (7n - 3)(3n + 2)=0$ [Group Again]

Now, If :

$3 n + 2 = 0$ $3n+2=0$ , then $3 n = - 2 \Rightarrow n = - \frac{2}{3}$ $3n=-2 rArr n= -2/3$

And If :

$7 n - 3 = 0$ $7n-3=0$ , then $7 n = 3 \Rightarrow n = \frac{3}{7}$ $7n=3 rArr n = 3/7$

Hence Explained.

Answer 2 · 2018-03-28T13:03:03

$n = - \frac{2}{3} or n = \frac{3}{7}$ $n=-2/3" or "n=3/7$

Explanation:

$factorise the quadratic using the a-c method$ $"factorise the quadratic using the a-c method"$

$that is factors of 21 \times - 6 = - 126 which sum to + 5$ $"that is factors of "21xx-6=-126" which sum to "+5$

$the factors of - 126 which sum to + 5 are + 14 and - 9$ $"the factors of "-126" which sum to + 5 are + 14 and - 9"$

$split the middle term using these factors$ $color(blue)"split the middle term using these factors"$

$\Rightarrow 21 n^{2} + 14 n - 9 n - 6 = 0 \leftarrow factor by grouping$ $rArr21n^2+14n-9n-6=0larrcolor(blue)"factor by grouping"$

$\Rightarrow 7 n (3 n + 2) - 3 (3 n + 2) = 0$ $rArrcolor(red)(7n)(3n+2)color(red)(-3)(3n+2)=0$

$take out the common factor (3 n + 2)$ $"take out the common factor "(3n+2)$

$\Rightarrow (3 n + 2) (7 n - 3) = 0$ $rArr(3n+2)(color(red)(7n-3))=0$

$equate each factor to zero and solve for n$ $"equate each factor to zero and solve for n"$

$3 n + 2 = 0 \Rightarrow n = - \frac{2}{3}$ $3n+2=0rArrn=-2/3$

$7 n - 3 = 0 \Rightarrow n = \frac{3}{7}$ $7n-3=0rArrn=3/7$

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How do you solve $21 n^{2} + 5 n - 6 = 0$ $21n ^ { 2} + 5n - 6= 0$ ?

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