Arithmetic · real student question

Evaluate (1/2) to the zero power, plus [(3/2 - 1/4) squared minus (1 - 1/4) squared] divided by (3/2) squared, minus 1/2.

Question

Evaluate (12)0+[(32−14)2−(1−14)2]:(32)2−(12)1.\left(\frac12\right)^0+\left[\left(\frac32-\frac14\right)^2-\left(1-\frac14\right)^2\right]:\left(\frac32\right)^2-\left(\frac12\right)^1.

Step-by-step solution

  1. Resolve the two easy powers first. (12)0=1\left(\tfrac12\right)^0 = 1 (anything nonzero to the zero power) and (12)1=12\left(\tfrac12\right)^1 = \tfrac12.

  2. Work inside the parentheses before squaring. 32−14=64−14=54,1−14=34.\frac32-\frac14 = \frac64-\frac14 = \frac54, \qquad 1-\frac14 = \frac34. Squaring before subtracting - a common slip - would give a different and wrong value.

  3. Square and subtract inside the bracket. (54)2−(34)2=2516−916=1616=1.\left(\frac54\right)^2-\left(\frac34\right)^2 = \frac{25}{16}-\frac{9}{16} = \frac{16}{16} = 1. The difference-of-squares shortcut confirms it: (54−34)(54+34)=12⋅2=1\left(\tfrac54-\tfrac34\right)\left(\tfrac54+\tfrac34\right) = \tfrac12\cdot 2 = 1.

  4. Carry out the division. (32)2=94\left(\tfrac32\right)^2 = \tfrac94, so 1:94=49.1:\frac94 = \frac49.

  5. Assemble the three surviving terms. 1+49−12.1+\frac49-\frac12.

  6. Add over the common denominator 18. 1=18181 = \tfrac{18}{18}, 49=818\tfrac49 = \tfrac{8}{18}, 12=918\tfrac12 = \tfrac{9}{18}, so 18+8−918=1718.\frac{18+8-9}{18} = \frac{17}{18}.

Answer

1718\frac{17}{18}

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