anonymous
  • anonymous
Rationalize the denominator and simplify.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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AbdullahM
  • AbdullahM
Hi, welcome to OpenStudy.
anonymous
  • anonymous
thanks

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AbdullahM
  • AbdullahM
To rationalize the denominator, you must multiply the numerator and denominator by the conjugate of the denominator. What is the conjugate of \(\sqrt{a}-2\sqrt{y}\)?
anonymous
  • anonymous
you would just switch the subtraction to addition right?
anonymous
  • anonymous
how did you make that square root symbol
AbdullahM
  • AbdullahM
It's called latex. `\(\sqrt{a}\)` That's the code to do it. You can also use the equation button to do it easier :)
anonymous
  • anonymous
ok cool so wouldnt it be
anonymous
  • anonymous
\[\sqrt{a}+2\sqrt{y}\]
AbdullahM
  • AbdullahM
Correct!
AbdullahM
  • AbdullahM
So now, lets do the multiplication. \(\Large\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y}}\times \frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}+2\sqrt{y}}=?\)
AbdullahM
  • AbdullahM
Hint: \(\sf\Large (x+y)(x+y)=x^2+2xy+y^2\) \(\sf\Large (x+y)(x-y)=x^2-y^2\)
anonymous
  • anonymous
you lost me
AbdullahM
  • AbdullahM
Ok, lets multiply the numerators first. \(\sf\Large (\sqrt{a}+2\sqrt{y})\times (\sqrt{a}+2\sqrt{y})=?\)
AbdullahM
  • AbdullahM
Use the formula I gave you above.
anonymous
  • anonymous
\[\sqrt{a^2}+4\sqrt{y^2}\]
anonymous
  • anonymous
?
AbdullahM
  • AbdullahM
No...
AbdullahM
  • AbdullahM
\(\sf\Large (\sqrt{a}+2\sqrt{y})\times (\sqrt{a}+2\sqrt{y})=?\) Use the formula: \(\sf\Large (x+y)\times (x+y)= x^2+2xy+y^2\) remember: x, y, a, and b are all variables.
anonymous
  • anonymous
im lost
anonymous
  • anonymous
@Lyralei can you help me
AbdullahM
  • AbdullahM
Where are you lost at?

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