1.solve for g. √6g - 23 = √12 - g
2.solve for d. √d^2 -11 = 5

Question

a1234 · Answer

I think you need to square each side and then solve it like a regular equation.

a1234 · Answer

That is for the first equation. I don't know about the second one.

kirbykirby · Answer

Is the equation $\sqrt{d^2-11}=5$ or $\sqrt{d^2}-11=5$

anonymous · Answer

its the first one

kirbykirby · Answer

$\sqrt{d^2-11}=5$
$\left(\sqrt{d^2-11}\right)^2=5^2$
$d^2-11=25$
$d^2=36$
$d=\pm \sqrt{36}=\pm 6$

anonymous · Answer

what about the second one??

kirbykirby · Answer

$\sqrt{6g - 23} = \sqrt{12 - g}$

$\left(\sqrt{6g - 23}\right)^2 = \left(\sqrt{12 - g}\right)^2$

$ 6g-23=12-g$

$7g=35$

$g=5$

anonymous · Answer

thanks could you help with a couple more

anonymous · Answer

which is the extraneous solution of -x = √2x + 15

kirbykirby · Answer

If you mean $\sqrt{2x+15}$ instead of $\sqrt{2x}+15$, please at least write √(2x+15) to say that you're taking the square root of all of those terms. Otherwise it is not clear.

anonymous · Answer

yes its the first one sorry

kirbykirby · Answer

$$-x=\sqrt{2x+15}\
(-x)^2=\left(\sqrt{2x+15}\right)^2\
x^2=2x+15\
x^2-2x-15=0\
(x-5)(x-3)=0\
x=5, \text{or }x=-3$$

But now look at the original equation...$-x=\sqrt{2x+15}$
Does it make sense that x=5? No! because that means $-5=\sqrt{2x+15}$
and a square root cannot equal a negative number.

But it works for x=-3, since you get $-(-3)=3=\sqrt{2x+15}$

So x=5 is the extraneous solution (meaning it is NOT a valid solution to the original equation)

kirbykirby · Answer

Oh I made a typo on the before last line.. it should read
$(x-5)(x+3)=0$

anonymous · Answer

thank you