Is it possible to simplify this radical equation without using a calculator?

[mathdad]

Show that the left side equals the right side without a calculator.

sqrt{7} - sqrt{8 - 2sqrt{7}} = 1

I know squaring must be done here and probably more than once. I am stuck in terms of squaring the left side.

[sqrt{7} - sqrt{8 - 2sqrt{7}}]^2 = (1)^2

Can someone square the left side for me step by step?

 

[topsquark]

Show that the left side equals the right side without a calculator.

sqrt{7} - sqrt{8 - 2sqrt{7}} = 1

I know squaring must be done here and probably more than once. I am stuck in terms of squaring the left side.

[sqrt{7} - sqrt{8 - 2sqrt{7}}]^2 = (1)^2

Can someone square the left side for me step by step?

You didn't do the square on the [math]\sqrt{7} - \sqrt{8 - 2 \sqrt{7}}[/math] right.

Let [math]a = \sqrt{7}[/math] and [math]b = \sqrt{8 - 2 \sqrt{7}}[/math]. What you wrote is [math](a - b)^2 = a^2 - b^2[/math] . This is not true. [math](a - b)^2 = a^2 - 2ab + b^2[/math] or:

[math]\left ( \sqrt{7} - \sqrt{8 - 2 \sqrt{7}} \right )^2 = (\sqrt{7})^2 - 2 \sqrt{7} \cdot \sqrt{8 - 2 \sqrt{7}} + \left ( \sqrt{8 - 2 \sqrt{7}} \right )^2[/math]

See what you can do with this.

-Dan

 

[mathdad]

You didn't do the square on the [math]\sqrt{7} - \sqrt{8 - 2 \sqrt{7}}[/math] right.

Let [math]a = \sqrt{7}[/math] and [math]b = \sqrt{8 - 2 \sqrt{7}}[/math]. What you wrote is [math](a - b)^2 = a^2 - b^2[/math] . This is not true. [math](a - b)^2 = a^2 - 2ab + b^2[/math] or:

[math]\left ( \sqrt{7} - \sqrt{8 - 2 \sqrt{7}} \right )^2 = (\sqrt{7})^2 - 2 \sqrt{7} \cdot \sqrt{8 - 2 \sqrt{7}} + \left ( \sqrt{8 - 2 \sqrt{7}} \right )^2[/math]

See what you can do with this.

-Dan

The middle part is still confusing.

2sqrt{7}*sqrt{8 - 2sqrt{7}}

It is legal to multiply sqrt{7} by sqrt{8 - 2sqrt{7}}?

In other words, can I apply the rule
sqrt{a}*sqrt{b} = sqrt{ab}?

I know the piece on the far right [sqrt{8 - 2sqrt{7}]^2 =
8 - 2sqrt{7}.

Can you break down the pieces for me if my effort is wrong?

 

[mathdad]

I used the wolfram website to calculate
(2sqrt{7})(sqrt{8 - 2sqrt{7}}) and several forms of the product were displayed.

I selected the form 2(7 - sqrt{7}) and was able to show that indeed the left side also equals 1.

(sqrt{7})^2 - 2(7 - sqrt{7}) + [sqrt{8 - 2sqrt{7}}]^2 = 1

(sqrt{7})^2 - 2(7 - sqrt{7}) + 8 - 2sqrt{7} = 1

7 - 14 + 2 sqrt{7}) + 8 - 2sqrt{7} = 1

7 - 14 + 8 = 1

15 - 14 = 1

1 = 1

Question:

How does (2sqrt{7})(sqrt{8 - 2sqrt{7}) become
2(7 - sqrt{7})?

 

[Greg]

$$\sqrt{7}-\sqrt{8-2\sqrt7}=\sqrt7-\sqrt{(\sqrt7-1)^2}=\sqrt7+1-\sqrt7=1$$

 

[mathdad]

$$\sqrt{7}-\sqrt{8-2\sqrt7}=\sqrt7-\sqrt{(\sqrt7-1)^2}=\sqrt7+1-\sqrt7=1$$

I had no idea that it is possible to bring sqrt{7} to the right side. I always thought that radicals and constants are to be separated when solving radical equations.