<img src="https://tex.z-dn.net/?f=x%20%2B%205%20%3D%200.5%28x%20%20%2B%203%29%20%5E%7B2%7D%20" id="TexFormula1" title="x + 5 = 0

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pashok25 [27]

2 years ago

.5(x + 3) ^{2} " alt="x + 5 = 0.5(x + 3) ^{2} " align="absmiddle" class="latex-formula">

Physics

1 answer:

nikitadnepr [17]2 years ago

6 0

Solve for x over the real numbers = complete the square
x + 5 = 0.5 (x + 3)^2

0.5 (x + 3)^2 = 1/2 (x + 3)^2:
x + 5 = 1/2 (x + 3)^2

Expand out terms of the right hand side:
x + 5 = x^2/2 + 3 x + 9/2

Subtract x^2/2 + 3 x + 9/2 from both sides:
-x^2/2 - 2 x + 1/2 = 0

Multiply both sides by -2:
x^2 + 4 x - 1 = 0

Add 1 to both sides:
x^2 + 4 x = 1

Add 4 to both sides:
x^2 + 4 x + 4 = 5

Write the left hand side as a square:
(x + 2)^2 = 5

Take the square root of both sides:
x + 2 = sqrt(5) or x + 2 = -sqrt(5)

Subtract 2 from both sides:
x = sqrt(5) - 2 or x + 2 = -sqrt(5)

Subtract 2 from both sides:
Answer: x = sqrt(5) - 2 or x = -2 - sqrt(5)

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