Question

help please (:


find the real root by graphing

29. x^2+3=x^3-5

Answer 1

well
\(\bf x^2+3=x^3-5\implies x^2+3-x^3+5=0
\\ \quad \\
-x^3+x^2+8=0\implies -x^3+x^2+8=y\impliedby \textit{graph that}\)

Answer 2

notice, 0 = y
that is, what are the values of "x", when y is 0, or what are the x-intercepts
or in quadratic lingo, what are the zeros and thus the solutions

Answer 3

are you mean to use a graphing calculator?
if so.. plug that in :)

if you need one, many out there available

Answer 4

just to avoid confusion =)
are you meant to use a graphing calculator? is what I meant =)