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

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In summary, the conversation discusses how to show that the left side of the equation sqrt{7} - sqrt{8 - 2sqrt{7}} = 1 equals the right side, without using a calculator. The method involves squaring both sides and simplifying the equation, with the steps being broken down and explained. The final result shows that the left side is equal to the right side, proving the equation to be true. The conversation also touches on the concept of multiplying radicals and constants, and how to properly simplify expressions involving them.
  • #1
mathdad
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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?
 
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  • #2
RTCNTC said:
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 \(\displaystyle \sqrt{7} - \sqrt{8 - 2 \sqrt{7}}\) right.

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

\(\displaystyle \left ( \sqrt{7} - \sqrt{8 - 2 \sqrt{7}} \right )^2 = (\sqrt{7})^2 - 2 \sqrt{7} \cdot \sqrt{8 - 2 \sqrt{7}} + \left ( \sqrt{8 - 2 \sqrt{7}} \right )^2\)

See what you can do with this.

-Dan
 
  • #3
topsquark said:
You didn't do the square on the \(\displaystyle \sqrt{7} - \sqrt{8 - 2 \sqrt{7}}\) right.

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

\(\displaystyle \left ( \sqrt{7} - \sqrt{8 - 2 \sqrt{7}} \right )^2 = (\sqrt{7})^2 - 2 \sqrt{7} \cdot \sqrt{8 - 2 \sqrt{7}} + \left ( \sqrt{8 - 2 \sqrt{7}} \right )^2\)

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?
 
  • #4
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})?
 
  • #5
$$\sqrt{7}-\sqrt{8-2\sqrt7}=\sqrt7-\sqrt{(\sqrt7-1)^2}=\sqrt7+1-\sqrt7=1$$
 
  • #6
greg1313 said:
$$\sqrt{7}-\sqrt{8-2\sqrt7}=\sqrt7-\sqrt{(\sqrt7-1)^2}=\sqrt7+1-\sqrt7=1$$
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.
 

1. What is a radical equation?

A radical equation is an equation that contains a variable within a radical, such as a square root or cube root.

2. How do I solve a radical equation?

To solve a radical equation, you must isolate the radical on one side of the equation and then square both sides to eliminate the radical. Be sure to check for extraneous solutions.

3. What is an extraneous solution?

An extraneous solution is a solution that satisfies the equation but does not satisfy the original problem or equation. These can occur when squaring both sides of a radical equation.

4. Can a radical equation have multiple solutions?

Yes, a radical equation can have multiple solutions. This is because when squaring both sides to eliminate the radical, it is possible to obtain both the positive and negative root.

5. Are there any restrictions when solving a radical equation?

Yes, there may be restrictions when solving a radical equation. For example, if the radical contains a variable in the denominator, the value of the variable cannot make the denominator equal to zero.

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