Question #4a68b

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Answer 1 · 2016-10-19T07:06:43

$x = 15/4$

Note that as we have $4-x$ under a radical, we must have $x<=4$ to avoid taking the root of a negative number.

$4+sqrt(10-x) = 6+sqrt(4-x)$

$=> sqrt(10-x) = 2+sqrt(4-x)$

$=> (sqrt(10-x))^2 = (2+sqrt(4-x))^2$

$=> 10-x = 2^2 + 2(2)sqrt(4-x) + (sqrt(4-x))^2$

$=> 10-x = 4 + 4sqrt(4-x) + 4 - x$

$=> 2 = 4sqrt(4-x)$

$=> sqrt(4-x) = 1/2$

$=> (sqrt(4-x))^2 = (1/2)^2$

$=> 4-x = 1/4$

$=> x = 4-1/4$

$:. x = 15/4$

Checking our result:

$4+sqrt(10-15/4) = 4+sqrt(25/4)$

$= 4+5/2$

$= 13/2$

$=6+1/2$

$=6+sqrt(1/4)$

$=6+sqrt(4-15/4)$

as desired