// Numbas version: exam_results_page_options {"name": "Inbbavathie's copy of Complete the square and find solutions (SP1)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["big", "sml", "bits"], "rulesets": {}, "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.

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

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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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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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Now solve the quadratic equation

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

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Give the lowest solution first.

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$x_1=$ [[0]]

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or

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$x_2=$ [[1]]

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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}

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The coefficient of $x$ in the expanded quadratic.

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The constant term in the expanded quadratic.

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

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