Solution 2.3:6c
From Förberedande kurs i matematik 1
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- | {{ | + | If we complete the square of the expression, we have that |
- | + | ||
- | {{ | + | {{Displayed math||<math>\begin{align} |
+ | x^{2} - 5x + 7 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 7\\[5pt] | ||
+ | &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{28}{4}\\[5pt] | ||
+ | &= \Bigl(x-\frac{5}{2}\Bigr)^{2} + \frac{3}{4} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | and because <math>\bigl(x-\tfrac{5}{2}\bigr)^{2}</math> is a quadratic, this term assumes a minimal value zero when <math>x=5/2\,</math>. This shows that the polynomial's smallest value is <math>\tfrac{3}{4}</math>. |
Current revision
If we complete the square of the expression, we have that
\displaystyle \begin{align}
x^{2} - 5x + 7 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 7\\[5pt] &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{28}{4}\\[5pt] &= \Bigl(x-\frac{5}{2}\Bigr)^{2} + \frac{3}{4} \end{align} |
and because \displaystyle \bigl(x-\tfrac{5}{2}\bigr)^{2} is a quadratic, this term assumes a minimal value zero when \displaystyle x=5/2\,. This shows that the polynomial's smallest value is \displaystyle \tfrac{3}{4}.