# What is a if limit (x^2+x*`sqrt x` )=6a if x go to a?

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### 1 Answer

You need to evaluate the following limit, such that:

`lim_(x->a) (x^2 + xsqrt x) = a^2 + a*sqrt a`

The problem provides the result of the limit, hence, you need to equate the result you have obtained and the result provided by the problem, such that:

`a^2 + a*sqrt a = 6a => a^2 + a*sqrt a - 6a = 0`

Factoring out a yields:

`a(a + sqrt a - 6) = 0`

Using zero product rule yields:

`a = 0`

`a + sqrt a - 6 = 0 => sqrt a = 6 - a`

Squaring both sides yields:

`a = (6 - a)^2 => a = 36 - 12a + a^2 => a^2 - 13a + 36 = 0`

Using quadratic formula yields:

`a_(1,2) = (13+-sqrt(169 - 144))/2 => a_(1,2) = (13+-sqrt25)/2`

`a_1 = (13+5)/2 => a_1 = 9`

`a_2 = (13-5)/2 => a_2 = 4`

Testing the obtained values in limit yields:

`a = 0 => lim_(x->0) (0^2 + 0) = 6*0 = 0` valid

`a = 4 => lim_(x->4) (4^2 + 4sqrt 4) = 16 + 8 = 24 = 6*4` valid

`a = 9 => lim_(x->a) (9^2 + 9sqrt 9) = 81 + 27 = 108 != 6*9`

**Hence, evaluating the valud values of a, under the given conditions, yields **`a = 0, a = 4.`