because you squared the sqrt of x, this produced the extraneous root of 16. This is because, in general (-4)^2 can also produce 16, but the principle sqrt of 16 is +4 not -4. This is why one must always check solutions in the original equation because an interim solution (such as a solution of an interim quadratic) is not always a true solution.
because you squared the sqrt of x, this produced the extraneous root of 16. This is because, in general (-4)^2 can also produce 16, but the principle sqrt of 16 is +4 not -4. This is why one must always check solutions in the original equation because an interim solution (such as a solution of an interim quadratic) is not always a true solution.
{x+x ➖ }+{x+x ➖}={x^2+x^2 }=x^4 x^2^2 (x ➖ 2x+2).
9.
It's not complete we have Answer with i
Hi it's not complete
Зачем так запутано.
Y=√x
Y+Y^2=12
Y^2+Y-12=0
Дальше решаем квадратное уравнение и получаем
Y=3
Соответственно х=9
Y=3 or Y=-4 You cannot reject Y= -4 right away, maybe it will give you an imaginairy solution.
X= 9.
16 doesn't work in the original equation.