I am having trouble with my Grade 10 math homework (IM3+ IB system). The question is asking to find the intersecting point of two functions.
$y = 3/x$
and
$y = x^2 - 5x + 2$
our first step is to equate both 'x' sides to each other, resulting in "$3/x = x^2 - 5x + 2$"
by combining like terms this eventually turns into "$x^3 - 5x^2 + 2x - 3$" which is of course unfactorizeable, therefore I have to use the quadratic equation.
Although that is not possible. What am I supposed to do? Also, the answer is they intersect at point (4.71, 0.64)
Edit: I have the same question for another problem that intersects at point (-0.75, -0.43) and after combining like terms turns into "$-x^3 + x^2 - 1$"
If you can determine that the function $f(x)= x^3 - 5x^2 + 2x - 3$ is both convex and increasing, or concave and decreasing on an interval $[a,b]$ then that contains a solution of $f(x)=0$
Newton's method beginning at $b$ will converge to a solution. If it is concave and increasing, or convex and decreasing, and you start at $a$. For a cubic it's easy to identify the critical points and inflection and thus find such an interval.
However, you could try to factorise the above cubic using complex numbers. It can be proved that if a polynomial has real coefficients, then the non-real complex roots of such a polynomial, if there are any, must occur in these complex conjugate pairs. That's why there can't be such cubics with 1 or 3 non-real complex roots.
One other thing worth mentioning: by the Intermediate value theorem, all cubics with real coefficients must have at least 1 real root (giving another reason such a cubic can't have 3 non-real complex roots).