1 Answer

Ccgillett · Answer 1 · 2021-08-07T14:38:04+0000

Answer:

$a = - 1 - √(3) \: or \: a = - 1 + √(3)$

Explanation:

We want to solve

$3 {(a + 1)}^(2) + 2 = 11$

with two different methods.

Subtract 2 from both sides:

$3 {(a + 1)}^(2) = 11 - 2 \\ 3 {(a + 1)}^(2) = 9$

Divide through by 3

${(a + 1)}^(2) = 3$

Take square root:

$a + 1 = \pm √(3)$

$a = - 1\pm √(3)$

$a = - 1 - √(3) \: or \: a = - 1 + √(3)$

Method 2:

The given equation is

$3 {(a + 1)}^(2) + 2 = 11$

Expand

$3( {a}^(2) + 2a + 1) + 2 = 11$

Subtract 11

$3( {a}^(2) + 2a + 1) - 9 = 0$

Divide through by 3

${a}^(2) + 2a + 1 - 3 = 0 \\ {a}^(2) + 2a - 2 = 0$

Use the quadratic formula:

$a = \frac{ - 2 \pm \sqrt{ {2}^(2) - 4 * 1 * - 2} }{2 * 1}$

$a = ( - 2 \pm √( 12) )/(2)$

$a = ( - 2 \pm2 √(3) )/(2)$

$a = - 1 \pm√(3)$

$a = - 1 - √(3) \: or \: a = - 1 + √(3)$

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