Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you solve $\sqrt{8 - x} + 6 = 4$ $sqrt(8-x)+6=4$ ?

1 Answer

Aug 11, 2015

$\sqrt{8 - x} + 6 = 4$ $sqrt(8-x)+6 = 4$ has no Real solutions

Explanation:

$\sqrt{anything} \geq 0$ $sqrt("anything") >= 0$
$XXXX$ $color(white)("XXXX")$ (by definition/convention)

If $\sqrt{x - 8} + 6 = 4$ $sqrt(x-8)+6 = 4$
$XXXXXXXX$ $color(white)("XXXXXXXX")$ $\to$ $rarr$ $XXXX$ $color(white)("XXXX")$ $\sqrt{8 - x} = - 2 < 0$ $sqrt(8-x) = -2 < 0$

Related questions

See all questions in Radical Equations

Impact of this question

1398 views around the world

You can reuse this answer
Creative Commons License