How do you solve using the completing the square method $x^{2} - 8 x + 7 = 0$ $x^2-8x+7=0$ ?

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Answer 1 · 2016-05-06T15:02:56

The solutions are:
$x = 7, x = 1$ $color(green)(x = 7 , color(green)(x = 1$

$x^{2} - 8 x + 7 = 0$ $x^2 - 8x + 7 = 0$

$x^{2} - 8 x = - 7$ $x^2 - 8x = - 7$

To write the Left Hand Side as a Perfect Square, we add 16 to both sides
$x^{2} - 8 x + 16 = - 7 + 16$ $x^2 - 8x + 16 = - 7 + 16$

$x^{2} - 8 x + 16 = 9$ $x^2 - 8x + 16 = 9$

$x^{2} - 2 \cdot x \cdot 4 + 4^{2} = 9$ $x^2 - 2 * x * 4 + 4^2 = 9$

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

${(x - 4)}^{2} = 9$ $(x-4)^2 = 9$

$x - 4 = \sqrt{9}$ $x - 4 = sqrt9$ or $x - 4 = - \sqrt{9}$ $x -4 = -sqrt9$

$x - 4 = 3$ $x - 4 = 3$ or $x - 4 = - 3$ $x -4 = -3$

$x = 3 + 4$ $x = 3 + 4$ or $x = - 3 + 4$ $x = -3 + 4$

$x = 7$ $color(green)(x = 7$ or $x = 1$ $color(green)(x = 1$

How do you solve using the completing the square method x2−8x+7=0x^2-8x+7=0?