// Numbas version: exam_results_page_options {"name": "Solving a perfect square by factorising", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "b", "aa", "bb", "mid", "g", "gg", "c", "d"], "name": "Solving a perfect square by factorising", "tags": ["binomial", "factorisation", "Factorisation", "factorise", "perfect square", "quadratic", "quadratics", "solving", "square"], "preamble": {"css": "", "js": ""}, "advice": "", "rulesets": {}, "parts": [{"stepsPenalty": 0, "prompt": "

Solve the following quadratic by factorisation:

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 $\\simplify{{aa}x^2+{mid}x+{bb}}$ $=$ 0 $\\Longrightarrow$ [[0]] $=$ 0 $\\Longrightarrow$ $x$ $=$ [[1]]
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Note: In the first gap, enter the quadratic in factored form.

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Note: In the second gap, there should only be one solution.

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Ensure you factorise the expression.

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Ensure you factorise the expression.

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Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.

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In fact, $\\simplify{{aa}x^2+{mid}x+{bb}}$ is also a perfect square, since $\\var{aa}x^2=(\\var{a}x)^2$, $\\var{bb}=\\simplify{({b})^2}$, and $\\var{mid}x=2(\\var{a}x)(\\var{b})$.

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That is, $\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{({a}x+{b})^2}$.

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Now, since $\\simplify{({a}x+{b})^2}=0$ we can take the (plus or minus) square root of both sides to get $\\simplify{({a}x+{b})}=0$. We can then solve this for $x$ to find $x=\\simplify{-{b}/{a}}$.

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Each bracket has a common factor of $\\var{g}$, so we can move both of them to the front, to write

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$\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{{gg}({a/g}x+{b/g})^2}$.

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Now, using the null factor law, since $\\simplify{{gg}({a/g}x+{b/g})^2}=0$ we must have $\\simplify{{a/g}x+{b/g}}=0$. We can then solve this for $x$ to find $x=\\simplify{{-b}/{a}}$.

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Solve the following quadratic by factorisation:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ $=$ 0 $\\Longrightarrow$ [[0]] $=$ 0 $\\Longrightarrow$ $x$ $=$ [[1]]
\n

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Note: In the first gap, enter the quadratic in factored form.

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Note: In the second gap, there should only be one solution.

\n

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Ensure you factorise the expression.

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Ensure you factorise the expression.

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Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.

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In fact, $\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ is also a perfect square, since

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• $\\simplify{{c[0]^2}/{c[1]^2}}x^2=\\left(\\simplify{{c[0]}/{c[1]}}x\\right)^2$,
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• $\\simplify{{d^2}/{c[2]^2}}=\\left(\\simplify{{d}/{c[2]}}\\right)^2$ and
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• $\\simplify{{2c[0]*d}/{(c[1]*c[2])}x}=2\\left(\\simplify{{c[0]}/{c[1]}x}\\right)\\left(\\simplify{{d}/{c[2]}}\\right)$.
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That is,

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$\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}=\\simplify{({c[0]}/{c[1]}x+{d}/{c[2]})^2}$.

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Now, since $\\simplify{({c[0]}/{c[1]}x+{d}/{c[2]})^2}=0$ we can take the (plus or minus) square root of both sides to get $\\simplify{{c[0]}/{c[1]}x+{d}/{c[2]}}=0$. We can then solve this for $x$ to find $x=\\simplify{{-d*c[1]}/{c[0]*c[2]}}$.

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