Solution 2.3:5b

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


\displaystyle \begin{align} & 4x^{2}-28x+48=4\left( x^{2}-7x+12 \right)=4\left( \left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+12 \right) \\ & =4\left( \left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{48}{4} \right)=4\left( \left( x-\frac{7}{2} \right)^{2}-\frac{1}{4} \right)=4\left( x-\frac{7}{2} \right)^{2}-1. \\ \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 \displaystyle -\text{1}.

NOTE: All values of \displaystyle x between \displaystyle \text{3} and \displaystyle \text{4} give a negative value for the expression.