How do you solve $7 a^{2} + 24 = - 29 a$ $7a^2+24=-29a$ ?

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How do you solve $7 a^{2} + 24 = - 29 a$ $7a^2+24=-29a$ ?

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Answer 1 · 2015-06-18T12:09:59

I solve by writing as a quadratic, then factoring. (If factoring hadn't worked quickly, I would have used the quadratic formula.)

Explanation:

$7 a^{2} + 24 = - 29 a$ $7a^2+24=-29a$ if and only if

$7 a^{2} + 29 a + 24 = 0$ $7a^2 + 29a+24 = 0$

If it is easily factorable, the factors must look like:

$(7 a + sss) (a + sss)$ $(7a + color(white)"sss" )(a + color(white)"sss")$

And the spaces are filled by two factors of 24:

$(7 a + m) (a + n)$ $(7a+m)(a+n)$

$m \times n = 24$ $m xx n = 24$ where $m a + 7 n a = 29 a$ $ma+7na = 29a$

$1 \times 24$ $1 xx 24$ and $24 \times 1$ $24 xx 1$ won't work
$2 \times 12$ $2 xx 12$ and $12 \times 2$ $12 xx 2$ won't work
$3 \times 8$ $3 xx 8$ won't work in that order, but $8 \times 3$ $8 xx 3$ works.

Check to be sure
$(7 a + 8) (a + 3) = 7 a^{2} + 21 a + 8 a + 24 = 7 a^{2} + 29 a + 24$ $(7a+8)(a+3) = 7a^2 +21a+8a+24 = 7a^2 +29a +24$

So we have:
$7 a^{2} + 29 a + 24 = 0$ $7a^2 + 29a+24 = 0$

$(7 a + 8) (a + 3) = 0$ $(7a+8)(a+3) = 0$

$7 a + 8 = 0$ $7a+8 =0$ $or$ $" or "$ $a + 3 = 0$ $a+3=0$

$a = - \frac{8}{7}$ $a = -8/7$ $or$ $" or "$ $a = - 3$ $a = -3$

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How do you solve $7 a^{2} + 24 = - 29 a$ $7a^2+24=-29a$ ?

1 Answer

Explanation:

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How do you solve 7a^2+24=-29a7a2+24=−29a7a^2+24=-29a?

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Explanation:

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How do you solve $7 a^{2} + 24 = - 29 a$ $7a^2+24=-29a$ ?