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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} ", "variables": {"bits": {"description": " After completing the square, the expression will have the form (x + \\mathrm{bits}[0])^2 - \\mathrm{bits}[1]^2. ", "definition": "sort(shuffle(1..9)[0..2])", "group": "Ungrouped variables", "name": "bits", "templateType": "anything"}, "big": {"description": " The constant term in the expanded quadratic. ", "definition": "bits[0]^2-bits[1]^2", "group": "Ungrouped variables", "name": "big", "templateType": "anything"}, "sml": {"description": " The coefficient of x in the expanded quadratic. ", "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\". \n This can be useful when it isn't obvious how to fully factorise a quadratic equation. ", "tags": [], "name": "Complete the square and find solutions", "ungrouped_variables": ["big", "sml", "bits"], "variable_groups": [], "parts": [{"showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "scripts": {}, "prompt": " Write the following expression in the form a(x+b)^2-c. \n \\simplify {x^2+{sml}x+{big}} =  [[0]] ", "type": "gapfill", "unitTests": [], "variableReplacements": [], "marks": 0, "gaps": [{"showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "vsetRange": [0, 1], "failureRate": 1, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "showPreview": true, "musthave": {"message": " It doesn't look like you've completed the square. ", "strings": [")^2"], "partialCredit": 0, "showStrings": false}, "checkVariableNames": false, "type": "jme", "expectedVariableNames": [], "unitTests": [], "variableReplacements": [], "marks": 1, "checkingAccuracy": 0.001, "scripts": {}, "answer": "(x+{bits[0]})^2-{bits[1]^2}", "vsetRangePoints": 5, "notallowed": {"message": " It doesn't look like you've completed the square. ", "strings": ["x^2"], "partialCredit": 0, "showStrings": false}, "checkingType": "absdiff"}], "sortAnswers": false}, {"showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "scripts": {}, "prompt": " Now solve the quadratic equation \n \\[ \\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]]

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

"}, "type": "question", "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Johnny Yi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2810/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Johnny Yi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2810/"}]}