Question

Algebra 2 help

Answer 1

\[ T>0, then \sqrt{\frac{ (T^2+T)(T^3+T^2) }{ (T^2+T)+(T^3+T^2) }}\]

Answer 2

=T

Answer 3

\[\sqrt{\frac{ (T^{2}+T)(T^{2}+T^{2}) }{ (T^{2}+T)+(T^{2}+T) }}\]
\[\sqrt{\frac{ (T^{2}+T)(2T^{2}) }{ 2(T^{2}+T) }}\]
\[\sqrt{\frac{ 2T^{2} }{ 2 }}\]
\[\sqrt{T^{2}}\]
T

Answer 4

Multiply the numerator. Then separate the HCF from numerator and numerator. You will find that what is left is the same, so you will cancel it out. Then cancel out what you can of the HCF on numerator and denominator, and you will be left with t squared. So the answer is T

Answer 5

@shrinifores You got the right answer, but you actually did a different problem! Your numerator and denominators are different than the ones in the problem statement.

\[\sqrt{\frac{(T^2+T)(T^3+T^2)}{(T^2+T)+(T^3+T^2)}}\]Expand the numerator, then factor:
\[\sqrt{\frac{T^5+2T^4+T^3}{(T^2+T)+(T^3+T^2)}} = \sqrt{\frac{T^3(T+1)^2}{(T^2+T)+(T^3+T^2)}}\]Expand the denominator, factor and simplify.
\[\sqrt{\frac{T^3(T+1)^2}{(T^3+2T^2+T)}} = \sqrt{\frac{T^3(T+1)^2}{T(T+1)^2}} = \sqrt{T^2} = T\]Last step only possible because T known to be positive.

Answer 6

@whpalmer4 , thanks! Can you show me how you expanded the denominator on the next to last step? I dont understand that.

Answer 7

Isn't it already solved?

Answer 8

No, I have that last question

Answer 9

I want to know how he expanded the denominator

Answer 10

\[(T^2+T)+(T^3+T^2) = T^3 + 2T^2+T = T(T^2+2T+1) = T(T+1)(T+1)\]

Answer 11

Just adding the two expressions, factoring out a T, then recognizing that \(T^2+2T+1 = (T+1)^2\)

Answer 12

That's one of the common patterns which are helpful to recognize:

\((a+b)^2 = a^2+2ab + b^2\)
\((a-b)^2 = a^2-b^2\)

Answer 13

Thank you

Answer 14

the other way to treat the denominator:
\[ (T^2+T)+(T^3+T^2)\]
factor a T out of the first term, and T^2 out of the 2nd term
\[T(T+1) + T^2(T+1) \]
now factor (T+1) out of both terms
\[ (T+T^2)(T+1) \\ T(1+T)(T+1) \\ T(T+1)^2 \]

Answer 15

Oh @whpalmer4 i'm so sry...