How do you evaluate ${(\frac{- 3}{2})}^{2} + 3 (\frac{- 3}{2}) + 2$ $(\frac{-3}{2})^{2}+3(\frac{-3}{2})+2$ ?

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Answer 1 · 2017-11-05T14:49:58

Let's first deal with the indices and brackets.

${(- \frac{3}{2})}^{2} = \frac{9}{4}$ $(-3/2)^2 = 9/4$

$3 (- \frac{3}{2}) = - \frac{9}{2}$ $3(-3/2) = -9/2$

Now the problem looks like this:

$\frac{9}{4} - \frac{9}{2} + 2$ $9/4 -9/2 +2$

Make the fractions have a common denominator of 4 and then work it out

$\frac{8}{4} = 2$ $8/4 = 2$

$\frac{9}{4} - \frac{18}{4} + \frac{8}{4} \to \frac{9 - 18 + 8}{4}$ $9/4 - 18/4 + 8/4 -> (9-18+8)/4$

$= - \frac{1}{4}$ $=-1/4$

How do you evaluate (−32)2+3(−32)+2(\frac{-3}{2})^{2}+3(\frac{-3}{2})+2?