How do you solve $a^{2} - 4 a - 21 = 0$ $a^2 - 4a - 21 = 0$ by completing the square?

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Answer 1 · 2016-05-24T15:25:59

The solutions are:
$a = 7$ $color(green)(a = 7$ , $a = - 3$ $color(green)(a = -3$

$a^{2} - 4 a - 21 = 0$ $a^2 -4a - 21 = 0$

$a^{2} - 4 a = 21$ $a^2 -4a = 21$

To write the Left Hand Side as a Perfect Square, we add 4 to both sides
$a^{2} - 4 a + 4 = 21 + 4$ $a^2 -4a + color(green)(4) = 21 + color(green)(4$

$a^{2} - 2 \cdot a \cdot 2 + 2^{2} = 25$ $a^2 - 2 * a * 2 + 2^2 = 25$

Using the Identity ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $color(blue)((a-b)^2 = a^2 - 2ab + b^2$ , we get

${(a - 2)}^{2} = 25$ $(a-2)^2 = 25$

$a - 2 = \sqrt{25}$ $a-2 = sqrt25$ or $a - 2 = - \sqrt{25}$ $a-2 = -sqrt25$

$a - 2 = 5$ $a-2 = 5$ or $a - 2 = - 5$ $a-2 = -5$

$a = 5 + 2$ $a = 5+2$ or $a = - 5 + 2$ $a = -5+2$

$a = 7$ $color(green)(a = 7$ , $a = - 3$ $color(green)(a = -3$

How do you solve a^2 - 4a - 21 = 0a2−4a−21=0 a^2 - 4a - 21 = 0 by completing the square?