Completing the square works by noticing that
\n\\[ (x+a)^2 = x^2 + 2ax + a^2 \\]
\nSo when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.
\nRewrite $x^2+\\var{sml}x$ as $\\simplify[basic]{ (x+{sml/2})^2 - {sml/2}^2}$.
\n\\begin{align}
\\simplify[basic]{x^2+{sml}x+{big}} &= \\simplify[basic]{(x+{sml/2})^2-{(sml/2)}^2+{big}} \\\\
&= \\simplify[basic]{(x+{sml/2})^2+{-(sml/2)^2+big}} \\text{.}
\\end{align}
We showed above that
\n\\[ \\simplify[basic]{x^2+{sml}x+{big}} = 0 \\]
\nis equivalent to
\n\\[ \\simplify[basic]{(x+{bits[0]})^2-{bits[1]^2}} = 0 \\text{.} \\]
\nWe can then rearrange this equation to solve for $x$.
\n\\begin{align}
\\simplify{(x+{bits[0]})^2-{(bits[1])^2} } &= 0 \\\\
(x+\\var{bits[0]})^2 &= \\var{bits[1]^2} \\\\
x+\\var{bits[0]} &= \\pm \\var{bits[1]} \\\\
x &= -\\var{bits[0]} \\pm \\var{bits[1]} \\\\[2em]
x_1 &= \\var{-bits[0]-bits[1]} \\text{,}\\\\
x_2 &= \\var{-bits[0]+bits[1]} \\text{.}
\\end{align}
After completing the square, the expression will have the form $(x + \\mathrm{bits}[0])^2 - \\mathrm{bits}[1]^2$.
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", "definition": "2*bits[0]", "group": "Ungrouped variables", "name": "sml", "templateType": "anything"}}, "statement": "We can rewrite quadratic equations given in the form $ax^2+bx+c$ as a square plus another term - this is called \"completing the square\".
\nThis can be useful when it isn't obvious how to fully factorise a quadratic equation.
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\n$x_1=$ [[0]]
\nor
\n$x_2=$ [[1]]
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