I am to solve for x using square root property:
$3(x-4)^2=15$
The textbook solution is $4+-\sqrt{5}$ and I am unable to arrive at this. I arrived at $\sqrt{21}$
Here is my working:
$3(x-4)^2=15$
$(x-4)^2=5$ # divide both sides by factor 3
$x^2-16=5$ # multiply out $(x-4)^2$
$x^2 = 21$
$x=\sqrt{21}$
How can I arrive at $x=4+-\sqrt{5}$ # not sure syntax for 'plus or minus' symbol here, just used +-
$\endgroup$ 22 Answers
$\begingroup$We have$(x-4)^2 = 5$
Taking sqrt on both sides,
$$x-4 = \pm\sqrt5$$$$x = 4 \pm\sqrt5$$That's it!
$\endgroup$ 8 $\begingroup$Simplifying we get$$(x-4)^2=5$$ and expanding $$x^2-8x+11=0$$ so $$x_{1,2}=4\pm \sqrt{16-11}$$
$\endgroup$ 4
