Question

Jeremiah is factoring 0.64t^24 – 9 by using the rule a^2 – b^2 = (a + b)(a – b).

What will he use for the value of a?

A. 0.8t^12

B. 0.08t^12

C. 0.8

D. 0.64

Answer 1

`0.8\, t^(12)` (assuming that
`a \ge 0`.)

Explanation:

The rule
`a^(2) - b^(2) = (a + b)\, (a - b)` describes a way for factoring the difference between two perfect squares
`a^(2)` and
`b^(2)`.

Setting
`a^(2) = 0.64\, t^(24)` and
`b^(2) = 9` would ensure that
`(a^(2) - b^(2))` match the given expression.

Take the square root of both sides of the equality
`a^(2) = 0.64\, t^(24)`. Make use of the fact that for all
`x \ge 0`,
`√(x) = x^(1/2)`.

`\displaystyle \sqrt{a^(2)} = \sqrt{0.64\, t^(24)}`.

`\begin{aligned} a &= \left(√(0.64)\right)\, \sqrt{t^(24)} \\ &= 0.8\, (t^(24))^(1/2)\\ &= 0.8 \, t^(24 * (1/2)) \\ &= 0.8\, t^(12)\end{aligned}`.