anonymous
  • anonymous
I don't understand the underlined step
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Please open the attachment for details
1 Attachment
anonymous
  • anonymous
We are given that (a+b) = (c+d). If that is true, then (a+b)^2 must equal (c+d)^2. But that means that \(a + 2\sqrt{ab} + b = c + 2\sqrt{cd} + d\) And since a+b = c+d we can subtract those terms from both side to get that \[\sqrt{ab} = \sqrt{cd}\] Now we look at: \[(\sqrt{a}-\sqrt{b})^2 = a -2\sqrt{ab} + b\] \[(\sqrt{c}-\sqrt{d})^2 = c -2\sqrt{cd} + d\] But here again, a+b = c+d, and \(\sqrt{ab} = \sqrt{cd}\) So we can see that the two equations are equal and therefore \[(\sqrt{a}-\sqrt{b})^2 =(\sqrt{c}-\sqrt{d})^2\]
anonymous
  • anonymous
Ack, sorry I missed something in the first step.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
We are given that \(\sqrt{a} + \sqrt{b} = \sqrt{c} + \sqrt{d}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.