Question

Help needed plezAns in (ii) is 1.5

Answer 1

OK first let's check the x=1.5. 

$y = \frac x 6 (x^2 - 10)$

$y = 1-2x$

$1-2x = \frac x 6 (x^2 - 10)$

$6-12x = x^3 - 10 x$

$x^3 + 2x-6 = 0$

Oh my, that's called a depressed cubic, no $x^2$ term. There's a formula for these very much like the quadratic formula but you're probably not quite old enough for that.  Anyway, $x\approx 1.45616 \approx 1.5$ is a solution, but that's not what they're asking.  They are asking us to compare 

$x^3 + 2x-6 = 0$

with

$x^3 + bx^2 + cx + d = 0$

and conclude $b=0, c= 2, d=-6$

It turns out we did need all the rest of it.  Save those brain cells, there's lots more math coming.


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I love it when the student asks for more.  Here's the formula for a depressed cubic. I won't derive it here (though I did earlier today, coincidentally, but I'm probably not allowed to link to my Quora answer "what led to the discovery of complex numbers" from here).  We use the trick of putting coefficients on the coefficients to avoid fractions.

$x^3 + 3 p x = 2 q$

has solutions

$x = \sqrt[3] { q - \sqrt{p^3 + q^2} } + \sqrt[3] {q + \sqrt{p^3 + q^2} }$

That's pretty simple, though sometimes we end up having to take the cube roots of complex numbers, which isn't that helpful.  Let's try it out on

$x^3 + 2x=6$

That's $p=2/3, q=3$ so

$x = \sqrt[3] { 3 - \sqrt{(2/3)^3+9} } + \sqrt[3] {3 + \sqrt{(2/3)^3+9} }$

$x = \sqrt[3] { 3 - \sqrt{753}/9 } +\sqrt[3]{3 + \sqrt{753}/9 }$

$x \approx 1.4561642461359084609748069666$