Determining Equations with Squared Roots: Learn from O, P, and Q | Expert Help"

[sissi169]

if o,p,q are roots of the equation ax^3+bx^2+cx+d=0, determine the equation whose roots are o^2,p^2 and q^2

who can help me solve it? thank you

 

[malawi_glenn]

This is missplaced.

Please study the rules of this forum before posting.

 

[berkeman]

Thread moved to Homework Help, Precalc Math.

 

[Tedjn]

Unless I am missing some completely obvious substitution (and I may be), this question is solvable, but not trivial to solve or explain. I'm assuming you want a new function that only has o², p², and q² as roots; thus it will be cubic (barring imaginary roots). Only the constants a, b, c, and d will be different, and we want to write the new constants in terms of a, b, c, and d.

What you will need to do is to realize that $ax^3+bx^2+cx+d$ and $x^3+(b/a)x^2+(c/a)x + (d/a)$ have the same roots, because the second one is just the first one divided by $a \neq 0$. The right hand side is still 0, so the solutions aren't changed.

Now, we know we can factor and get

$$x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} = (x-o)(x-p)(x-q)$$

Multiply out and find b/a, c/a, and d/a in terms of o, p, and q. Then, suppose that the new cubic equation we want, with roots o², p², and q² is

$$x^3 + ex^2 + fx + g = (x-o^2)(x-p^2)(x-q^2)$$

Find e, f, and g in terms of o², p², and q². Try to relate it to b/a, c/a, and d/a.

That last part takes some thought, but give it a try and post back if you're still stuck after trying awhile.

 

[rock.freak667]

Usually in Further math, if we have an equation,$ax^3+bx^2+cx+d=0$ whose roots are $\alpha, \beta, \gamma$ and we want to find an equation whose roots are $\alpha^2,\beta^2, \gamma^2$. A substitution such as $y=x^2$ gives you it directly.

 

[Tedjn]

Except what happens if the roots are negative? That's the problem I had with substituting y = x². You will end up with a new equation, but the square root is always positive, so you may not come up with the same solutions. Maybe I am getting myself confused; your idea seems the more reasonable answer.

 

[rock.freak667]

sub $x=\sqrt{y}$ into the equation and simplify. It doesn't matter whether or not the roots are -ve or +ve.

 

[Tedjn]

Suppose f(x) = (x + 1)³. The triple root is -1. If we substitute $\sqrt{y}$ we find $f(y) = (\sqrt{y} + 1)^3$ which doesn't have any real roots. Of course, this is only because we define the square root as positive, but I think that is a real concern.

 

[rock.freak667]

$f(x)=(x+1)^3$ has root $\alpha (=-1)$ for his equation, there are 3 not one.

 

[HallsofIvy]

If a polynomial, with leading coefficient a, has roots o, p, q, then it can be written as a(x-p)(x-q)(x-o)= ax³-a(p+q+o)x²+a (op+ oq+ pq)x- aopq= 0

Similarly, the equation whose roots are p², q², and o² must be of the form x³- (p²+ q²+ o²)x²- (o²p²+ o²q²+ p²q²)x- o²p²q²= 0.

You know -aopq= d so opq= d/a and o²p²q²= d²/a². That's the constant term of the equation you are seeking. In fact if you multiply out (o+ p+ q)² that will equal b²/a², that should give you most of what you want.