How do you multiply [a^(1/2) + 6^(1/2)] [a^(1/2) - 6^(1/2)]?

Apr 10, 2015

You could multiply out out using FOIL, or whatever idea you've been taught for multiplying binomials.

It's quicker to recognize that this product is of the form:

(x+y)(x-y) = x^2-y^2)

(The product of a sum and difference equals the difference of the squares.)

Now realize that ${\left({a}^{\frac{1}{2}}\right)}^{2} = a$ and ${\left({6}^{\frac{1}{2}}\right)}^{2} = 6$, so the product is:

$\left[{a}^{\frac{1}{2}} + {6}^{\frac{1}{2}}\right] \left[{a}^{\frac{1}{2}} - {6}^{\frac{1}{2}}\right] = {\left({a}^{\frac{1}{2}}\right)}^{2} - {\left({6}^{\frac{1}{2}}\right)}^{2} = a - 6$

OR

$\left[{a}^{\frac{1}{2}} + {6}^{\frac{1}{2}}\right] \left[{a}^{\frac{1}{2}} - {6}^{\frac{1}{2}}\right] = {a}^{\frac{1}{2}} {a}^{\frac{1}{2}} - {a}^{\frac{1}{2}} {6}^{\frac{1}{2}} + {a}^{\frac{1}{2}} {6}^{\frac{1}{2}} - {6}^{\frac{1}{2}} {6}^{\frac{1}{2}} = {a}^{\frac{1}{2} + \frac{1}{2}} - {6}^{\frac{1}{2} + \frac{1}{2}} = a - 6$