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In the proof of the Law of Cosines, the equation [tex]$c^2 = h^2 + (b-x)^2$[/tex] was created using the Pythagorean theorem. Which equation is a result of expanding [tex]$(b-x)^2$[/tex]?

A. [tex]c^2 = h^2 + b^2 - x^2[/tex]

B. [tex]c^2 = h^2 + b^2 - 2bx + x^2[/tex]

C. [tex]c^2 = h^2 + b^2 + x^2[/tex]

D. [tex]c^2 = h^2 + b^2 - 2bx - x^2[/tex]


Sagot :

To solve this problem, we need to expand the expression [tex]\((b - x)^2\)[/tex]. Let's start by recalling the algebraic formula for expanding a squared binomial.

The formula for expanding [tex]\((a - b)^2\)[/tex] is given by:
[tex]\[ (a - b)^2 = a^2 - 2ab + b^2 \][/tex]

In our problem, [tex]\(a = b\)[/tex] and [tex]\(b = x\)[/tex]. So, we'll use this formula to expand [tex]\((b - x)^2\)[/tex]:

[tex]\[ (b - x)^2 = b^2 - 2bx + x^2 \][/tex]

After expanding, we get:

[tex]\[ (b - x)^2 = b^2 - 2bx + x^2 \][/tex]

We now need to identify which given option matches this expanded form when incorporated into the equation [tex]\(c^2 = h^2 + (b - x)^2\)[/tex]:

1. [tex]\(c^2 = h^2 + b^2 - x^2\)[/tex]
2. [tex]\(c^2 = h^2 + b^2 - 2bx + x^2\)[/tex]
3. [tex]\(c^2 = h^2 + b^2 + x^2\)[/tex]
4. [tex]\(c^2 = h^2 + b^2 - 2bx - x^2\)[/tex]

Substituting our expanded form [tex]\((b - x)^2 = b^2 - 2bx + x^2\)[/tex] into [tex]\(c^2 = h^2 + (b - x)^2\)[/tex]:

[tex]\[ c^2 = h^2 + b^2 - 2bx + x^2 \][/tex]

Thus, the correct matching option is:

[tex]\[ c^2 = h^2 + b^2 - 2bx + x^2 \][/tex]

So, the equation that is the result of expanding [tex]\((b - x)^2\)[/tex] is:
[tex]\[ \boxed{c^2 = h^2 + b^2 - 2bx + x^2} \][/tex]
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