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This question takes the student through variety of examples of quadratic inequalities by asking them for the range(s) for which $x$ meets the inequality.

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a)

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This example is best illustrated by the graph of $\\simplify{f(x)=x^2+({a}-{b})x-{a}{b}}$ below. By finding the roots of the equation we can find the $x$ coordinates where the line crosses the $x$ axis and then we can use a sketch or visualise the graph to work out the set of values for $x$ where $f(x)>0$.\\[\\\\[0.1em]\\]

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{geogebra_applet('Hk5dTptY',[[\"a\",a],[\"b\",b]])}

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\\[
\\begin{align}
\\simplify{f(x)=x^2+({a}-{b})x-{a}{b}}&>0\\\\
\\simplify{(x+{a})(x-{b})}&=0\\text{.}
\\end{align}
\\]

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Therefore

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\\[ \\simplify{x<{-a}}\\text{ or }\\simplify{x>{b}}\\text{.}\\]

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b)

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We use the same method in this example but this time we use our graph to visualise where $g(x)<0$.\\[\\\\[0.1em]\\]

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{geogebra_applet('PUFStTNa',[[\"c\",c],[\"d\",d]])}

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\\[
\\begin{align}
\\simplify{f(x)=x^2+({c}-{d})x-{c}{d}}&<0\\\\
\\simplify{(x+{c})(x-{d})}&=0\\text{.}
\\end{align}
\\]

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Therefore

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\\[ \\simplify{x>{-c}}\\text{ and }\\simplify{x<{d}}\\text{.}\\]

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c)

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The notable difference with solving this equation is the requirement to rearrange the inequality before factorisation.\\[\\\\[0.1em]\\]

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{geogebra_applet('CZsSCdH6',[[\"e\",g],[\"f\",f]])}

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\\[
\\begin{align}
\\simplify{({g}+{f})x-{g}{f}}&>x^2\\\\
\\simplify{x^2-({g}+{f})x+{g}{f}}&<0\\\\
\\simplify{(x-{g})(x-{f})}&=0\\text{.}
\\end{align}
\\]

\n

Therefore

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\\[ \\simplify{x>{g}}\\text{ and }\\simplify{x<{f}}\\text{.}\\]

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Alternatively, we can plot the graph of $x^2$ against $\\simplify{({g}+{f})x-{g}{f}}$ to visualise the same result.

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{geogebra_applet('Ez697SNE',[[\"g\",g],[\"f\",f]])}

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From this graph we can see that the values of $x$ where $\\simplify{({g}+{f})x-{g}{f}}>x^2$ are the same as the values of $x$ where $\\simplify{x^2-({g}+{f})x+{g}{f}}<0$.

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Solve the following quadratic inequalities by firstly factorising $f(x)$ and then solving for $x$ when $f(x)=0$. It may be helpful to sketch each quadratic.

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Find the range of values for $x$ such that $\\simplify{f(x)=x^2+({a}-{b})x-{a}{b}}>0$.

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Factorise $f(x)$:

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$\\simplify{f(x)=x^2+({a}-{b})x-{a}{b}}=$ [[0]] $=0$.

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Hence,

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$x>$ [[1]] 

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[[3]]

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$x<$ [[2]]

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Find the range of values for $x$ such that $\\simplify{f(x)=x^2+({c}-{d})x-{c}{d}}<0$.

\n

Factorise $f(x)$:

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$\\simplify{f(x)=x^2+({c}-{d})x-{c}{d}}=$ [[0]] $=0$.

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Hence,

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$x<$ [[1]]

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[[3]] 

\n

$x>$ [[2]]

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Find the range of vaues such that: $\\simplify{({g}+{f})x-{g}{f}>x^2}$   

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Rearrange then factorise the inequality:

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[[0]] $<0$.

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Use the above result to find the range of values for $x$ such that $\\simplify{({g}+{f})x-{g}{f}>x^2}$.

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$x>$ [[1]] 

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[[3]]  

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$x<$ [[2]]

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