// Numbas version: exam_results_page_options {"name": "Complete the square and find solutions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"questions": [{"tags": ["taxonomy"], "contributors": [{"profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/", "name": "Christian Lawson-Perfect"}, {"profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/", "name": "Chris Graham"}], "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\".

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This can be useful when it isn't obvious how to fully factorise a quadratic equation.

", "functions": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "variables": {"bits": {"group": "Ungrouped variables", "description": "

After completing the square, the expression will have the form $(x + \\mathrm{bits}[0])^2 - \\mathrm{bits}[1]^2$.

", "name": "bits", "definition": "sort(shuffle(1..9)[0..2])", "templateType": "anything"}, "big": {"group": "Ungrouped variables", "description": "

The constant term in the expanded quadratic.

", "name": "big", "definition": "bits[0]^2-bits[1]^2", "templateType": "anything"}, "sml": {"group": "Ungrouped variables", "description": "

The coefficient of $x$ in the expanded quadratic.

", "name": "sml", "definition": "2*bits[0]", "templateType": "anything"}}, "rulesets": {}, "variable_groups": [], "name": "Complete the square and find solutions", "metadata": {"description": "

Solve a quadratic equation by completing the square. The roots are not pretty!

", "licence": "Creative Commons Attribution 4.0 International"}, "parts": [{"prompt": "

Write the following expression in the form $a(x+b)^2-c$.

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$\\simplify {x^2+{sml}x+{big}} =$ [[0]]

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It doesn't look like you've completed the square.

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It doesn't look like you've completed the square.

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\$\\simplify {x^2+{sml}x+{big}} = 0\\text{.} \$

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$x_1=$ [[0]]

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or

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$x_2=$ [[1]]

Completing the square works by noticing that

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\$(x+a)^2 = x^2 + 2ax + a^2 \$

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So when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.

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#### a)

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Rewrite $x^2+\\var{sml}x$ as $\\simplify[basic]{ (x+{sml/2})^2 - {sml/2}^2}$.

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\\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}

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#### b)

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We showed above that

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\$\\simplify[basic]{x^2+{sml}x+{big}} = 0 \$

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is equivalent to

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\$\\simplify[basic]{(x+{bits[0]})^2-{bits[1]^2}} = 0 \\text{.} \$

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We can then rearrange this equation to solve for $x$.

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\\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}

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