How to simplify these expressions: a) (2√8 + 5√12)² ; b) (-√7+ 2√7)(√7- 2√7)  c) (2√3-√5)(2√3+√5)

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A square root is simplified if (1) there are no perfect square factors in the radicand (2) There are no fractions in the radicand (3) there are no roots in the denominator.

(a) We begin by simplifying the roots:

`(2sqrt(8)+5sqrt(12))^2=(2sqrt(4*2)+5sqrt(4*3))^2`

`=(4sqrt(2)+10sqrt(3))^2`

Now we can expand the binomial: `(a+b)^2=a^2+2ab+b^2`

`(4sqrt(2)+10sqrt(3))^2=(4sqrt(2))^2+2(4sqrt(2))(10sqrt(3))+(10sqrt(3))^2`

`=16*2+80sqrt(6)+100*3`

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A square root is simplified if (1) there are no perfect square factors in the radicand (2) There are no fractions in the radicand (3) there are no roots in the denominator.

(a) We begin by simplifying the roots:

`(2sqrt(8)+5sqrt(12))^2=(2sqrt(4*2)+5sqrt(4*3))^2`

`=(4sqrt(2)+10sqrt(3))^2`

Now we can expand the binomial: `(a+b)^2=a^2+2ab+b^2`

`(4sqrt(2)+10sqrt(3))^2=(4sqrt(2))^2+2(4sqrt(2))(10sqrt(3))+(10sqrt(3))^2`

`=16*2+80sqrt(6)+100*3`

`=332+80sqrt(6)`

So the simplified form is `332+80sqrt(6)`

(b) `(-sqrt(7)+2sqrt(7))(sqrt(7)-2sqrt(7))`

First we can add like terms:

`-sqrt(7)+2sqrt(7)=sqrt(7)`  and `sqrt(7)-2sqrt(7)=-sqrt(7)`

So we have:

`(-sqrt(7)+2sqrt(7))(sqrt(7)-2sqrt(7))=(sqrt(7))(-sqrt(7))=-7`

So the simplified form is -7.

(c) `(2sqrt(3)-sqrt(5))(2sqrt(3)+sqrt(5))` ` `

` ` Here we can use the difference of two squares: `(a+b)(a-b)=a^2-b^2`

So `(2sqrt(3)-sqrt(5))(2sqrt(3)+sqrt(5))=(2sqrt(3))^2-(sqrt(5))^2`

`=(4*3)-5`

`=7`

So the answer is 7.

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