OpenStudy (anonymous):

Simplify (2 sqrt(5) + 3 sqrt(7))^2

3 years ago
OpenStudy (anonymous):

\[(2 \sqrt{5} + 3 \sqrt{7})^{2}\]

3 years ago
OpenStudy (mrnood):

(a+b)^2 = a^2 + 2ab + b^2 expand the brackets above using that formula

3 years ago
OpenStudy (mrnood):

be careful of brackets - it's (2sqrt(5))^2 for example

3 years ago
OpenStudy (mathstudent55):

@MrNood is doing a good job.

3 years ago
OpenStudy (mathstudent55):

You can use the square of a binomial pattern he showed you, or you can just use FOIL.

3 years ago
OpenStudy (mathstudent55):

You are close, but you need to be careful with the parentheses. @MrNood mentioned that too. \((2\sqrt5)^2 + 2 \times 2\sqrt5 \times 3\sqrt7 + (3\sqrt7)^2 \)

3 years ago
OpenStudy (anonymous):

ITS NOT CLOSED that was on accident

3 years ago
OpenStudy (anonymous):

I can still medal since i havent yet

3 years ago
OpenStudy (mathstudent55):

Ok, now you can square the first and last terms, and multiply out the middle term.

3 years ago
OpenStudy (anonymous):

How do i square the first and last terms?

3 years ago
OpenStudy (mathstudent55):

Square the number that multiplies the root. Then square the root, which is just the number without the root. Then you multiply them together. \((2\sqrt5)^2 + 12 \sqrt{35} + (3\sqrt7)^2\) \(= 4 \times 5 + 12\sqrt{35} + 9 \times 7\)

3 years ago
OpenStudy (anonymous):

iif i have to leave suddenly please dont leave ill be back on in like 3 minutes if it happens just stay if i do

3 years ago
OpenStudy (mathstudent55):

I already had a loss of connection, and soon I need to leave.

3 years ago
OpenStudy (mathstudent55):

@Darealest123

3 years ago
OpenStudy (anonymous):

So If you cant add it together how do you simplify, does it mean not to simplify all the way down to one number?

3 years ago
OpenStudy (mathstudent55):

The answer is: \(= 20 + 12\sqrt{35} + 63\) \(= 83 + 12\sqrt{35}\) That is the final answer.

3 years ago
OpenStudy (anonymous):

The second equation is the final answer? i need it in steps so i can write it so im trying to figure that out

3 years ago
OpenStudy (anonymous):

@mathstudent55

3 years ago
OpenStudy (anonymous):

nvm i figured it out here is the medal thank you so much

3 years ago
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