Solution 2.3:5b

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Instead of randomly trying different values of ''x'', it is better investigate the second-degree expression by completing the square,
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{{Displayed math||<math>\begin{align}
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4x^{2} - 28x + 48
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&= 4(x^{2} - 7x + 12)\\[5pt]
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&= 4\bigl((x-\tfrac{7}{2})^{2} - (\tfrac{7}{2})^{2} + 12\bigr)\\[5pt]
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&= 4\bigl((x-\tfrac{7}{2})^{2} - \tfrac{49}{4} + \tfrac{48}{4}\bigr)\\[5pt]
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&= 4\bigl((x-\tfrac{7}{2})^{2} - \tfrac{1}{4}\bigr)\\[5pt]
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&= 4\bigl(x - \tfrac{7}{2}\bigr)^{2}-1\,\textrm{.}
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\end{align}</math>}}
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In the expression in which the square has been completed, we see that if, e.g.
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<math>x=7/2</math>, then the whole expression is negative and equal to -1.
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Note: All values of ''x'' between 3 and 4 give a negative value for the expression.

Current revision

Instead of randomly trying different values of x, it is better investigate the second-degree expression by completing the square,

\displaystyle \begin{align}

4x^{2} - 28x + 48 &= 4(x^{2} - 7x + 12)\\[5pt] &= 4\bigl((x-\tfrac{7}{2})^{2} - (\tfrac{7}{2})^{2} + 12\bigr)\\[5pt] &= 4\bigl((x-\tfrac{7}{2})^{2} - \tfrac{49}{4} + \tfrac{48}{4}\bigr)\\[5pt] &= 4\bigl((x-\tfrac{7}{2})^{2} - \tfrac{1}{4}\bigr)\\[5pt] &= 4\bigl(x - \tfrac{7}{2}\bigr)^{2}-1\,\textrm{.} \end{align}

In the expression in which the square has been completed, we see that if, e.g. \displaystyle x=7/2, then the whole expression is negative and equal to -1.


Note: All values of x between 3 and 4 give a negative value for the expression.

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