Respuesta :

altavistard

Answer:

{1, 7}

Step-by-step explanation:

Start with x^2 - 8x + 7 = 0.

We want to modify the first two terms to resemble x^2 - ax + b - b, where a is the coefficient (here -8) of the x term and b is the square of half of a.

Here, a = -8.  Half of that is -4.  The square of -4 is 16, and this is the value of b.

Write out x^2 - 8x     as    x^2 - 8x + 16 - 16.  This is the perfect square of -4, added to x^2 - 8x and then subtracted from the result

Then the original equation, x^2 - 8x + 7 = 0, can be rewritten as:

x^2 - 8x + 16 - 16 + 7 = 0, and then as (x - 4)^2 - 9, or (x - 4)^2 = 9.

We want to solve this result for x.  To do this, take the square root of both sides:

x - 4 = ±√9, or x - 4 = ±3.

Adding 4 to both sides, we get:

x = 4 ± 3.  Thus, the solutions are x = 4 + 3 = 7 and x = 4 - 3 = 1.

BiologiaMagister

Answer:

[tex]x = 7, 1[/tex]

Step-by-step explanation:

[tex]x^{2} - 8x + 7 = 0[/tex]

[tex]x^{2} - 8x = -7[/tex]

[tex]x^{2} - 8x + (-4)^{2} = -7+(-4)^{2}[/tex] → the [tex]-4^{2}[/tex] was obtained by doing [tex]\frac{b}{2} ^{2}[/tex]

[tex]x^{2} - 8x + 16 = 9[/tex]

[tex](x - 4) = ± \sqrt{9}[/tex]

[tex](x - 4) = ± 3[/tex]

[tex]x = 3 + 4[/tex] or [tex]x = -3 + 4[/tex]

Note:

What makes this a "complete-the-square" procedure? - Well, we created a perfect suqare when doing [tex]\frac{b}{2}^{2}[/tex] , and that is what makes our procedure a "complete-the-suqare" procedure.

Hope it helped,

BioTeacher101