# Binomial question: What is the term that have x & y with the same power in `(root(3)(x^2)+sqrt y )^49` ?

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### 1 Answer

`(root(3)(x^2) + sqrty)^49`

Express the radicals as exponents.

`(x^(2/3) + y^(1/2))^49`

To determine the term in which the x and y have same exponents, use the formula in determining the nth term of the binomial `(a+b)^n` .

(r+1)th term `=``_nC_r a^(n-1)b^r`

Plug-in n=49 , `a=x^(2/3)` and `b=y^(1/2)` .

(r+1)th term `=` `_49 C_r (x^(2/3))^(49-r) y^(1/2)r`

(r+1)th term `=` `_49C_r x^(98/3-(2r)/3) y^(r/2)`

To solve for r, set the exponents of x and y equal.

`98/3 - (2r)/3 = r/2`

To simplify, multiply both sides by the LCD which is 6.

`6*(98/3 - (2r)/3) = r/2*6`

`196-4r=3r`

Isolate r. To do so, add both sides by 4r

`196+4r-4r=3r+4r`

`196=7r`

And divide both sides by 7.

`196/7=(7r)/7`

`28=r`

Next, plug-in r=28 to the nth term of `(x^(2/3)+y^(1/2))^49` .

(r+1)th term `=` `_49C_r x^(98/3-(2r)/3) y^(r/2)`

(28+1)the term`=` `_49C_28 x^(98/3-(2*28)/3)y^(28/2)`

29th term `=` `_49C_28 x^(98/3-56/3)y^14`

29th term `=` `_49C_28 x^(42/3)y^14`

29th term `=` `_49C_28 x^14y^14`

**Hence, it is the 29nth term of `(root(3)(x^2) + sqrty)^49` in which x and y **

**have same exponents. The term is `_49C_28 x^14 y^14` .**