Question #cabf2

Question

Question #cabf2

2 Answers

Impact of this question

1299 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-19T20:39:48

I found two real solutions:

Explanation:

I would try rearranging it as:
$x^2=4sqrt(x)$
square both sides:
$x^4=16x$
$x^4-16x=0$
$x(x^3-16)=0$
so we have:
$x=0$
and
$x^3-16=0$
$x^3=16$

solutions:
$x_1=0$
$x_2=root(3)(16)$

$x_3$ and $x_4$ should be two imaginary solutions considering that the graph of $x^3-16$ has only one real solution ( $x$ intercept):
graph{x^3-16 [-105.4, 105.4, -52.7, 52.8]}

Answer 2 · 2016-11-19T21:56:10

$x=2.52$

Explanation:

$x^2-4sqrtx=0$

$x^2=4sqrtx$

$x^2/sqrtx=x^(3/2)=4$

$x^(3/2)=(sqrtx)^3=4$

$sqrtx=root(3)4$

$x=(root(3)4)^2=2.52$

Science

Math

Humanities

... and beyond

Question #cabf2

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question