How do you solve $x^{2} + 16 x = 17$ $x^2+16x=17$ by completing the square?

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Answer 1 · 2016-04-13T16:16:01

$x = 1$ $color(green)(x = 1$ , $x = - 17$ $color(green)(x = -17$

$x^{2} + 16 x = 17$ $x^2 + 16x = 17$

To write the Left Hand Side as a Perfect Square, we add 64 to both sides:

$x^{2} + 16 x + 64 = 17 + 64$ $x^2 + 16x + 64 = 17 + 64$

$x^{2} + 2 \cdot x \cdot 8 + 8^{2} = 81$ $x^2 + 2 * x * 8 + 8^2 = 81$

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
${(x + 8)}^{2} = 81$ $(x+8)^2 = 81$

$x + 8 = \sqrt{81}$ $x + 8 = sqrt81$ or $x + 8 = - \sqrt{81}$ $x +8 = -sqrt81$

$x + 8 = 9$ $x + 8 = 9$ or $x + 8 = - 9$ $x +8 = -9$

$x = 9 - 8 = 1$ $x = 9 - 8 =1$ or $x = - 9 - 8$ $x = -9 - 8$

$x = 1$ $color(green)(x = 1$ or $x = - 17$ $color(green)(x = -17$

How do you solve x^2+16x=17x2+16x=17x^2+16x=17 by completing the square?