How do you solve $\frac{1}{2} {(x + 8)}^{2} = 14$ $1/2(x+8)^2=14$ ?

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How do you solve $\frac{1}{2} {(x + 8)}^{2} = 14$ $1/2(x+8)^2=14$ ?

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Answer 1 · 2017-01-28T00:13:33

$x = 2 \sqrt{7} - 8$ $x = 2sqrt(7) - 8$

Explanation:

$\frac{1}{2} {(x + 8)}^{2} = 14$ $1/"2"(x + 8)^2 = 14$

Multiplying by 2 on both sides

$cancel(2) × 1/cancel("2")(x + 8)^2 = 2 × 14$

$(x + 8)^2 = 28$

Square root both side

$sqrt((x + 8)^2$ = $sqrt(28)$

$x + 8 = +-2sqrt(7)$

$x = 2sqrt(7) - 8 and x =- 2sqrt(7) - 8$

Answer 2 · 2017-01-28T08:01:33

$x=-8+-2sqrt7$

Explanation:

$1/2 (x+8)^2=14$

I want to expand the squared term, so to do that I'll first multiply both sides by 2:

$1/2 color(red)(xx2) (x+8)^2=14color(red)(xx2)$

$(x+8)^2=28$

$x^2+16x+64=28$

$x^2+16x+64color(red)(-28)=28color(red)(-28)$

$x^2+16x+36=0$

I don't see any easy factors, so let's use the quadratic formula. The general formula is:

$x=(-color(green)b+-sqrt(color(green)b^2-4color(blue)acolor(brown)c))/(2color(blue)a)$

and relates to a trinomial this way:

$color(blue)ax^2+color(green)bx+color(brown)c$

$x=(-16+-sqrt(16^2-4(1)(36)))/(2(1))$

$x=(-16+-sqrt(256-144))/2$

$x=(-16+-sqrt(112))/2=(-16+-sqrt(16xx7))/2=(-16+-4sqrt(7))/2=-8+-2sqrt7$

And we can see these two solutions in the graphing of both sides of the original equation:

graph{(y-(1/2)(x+8)^2)(y-0x-14)=0 [-16.69, 3.31, 9.14, 19.136]}

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How do you solve $\frac{1}{2} {(x + 8)}^{2} = 14$ $1/2(x+8)^2=14$ ?

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How do you solve $\frac{1}{2} {(x + 8)}^{2} = 14$ $1/2(x+8)^2=14$ ?