MHB [ASK] Proof of Some Quadratic Functions

[Monoxdifly]

So, I found these statements and I need your assistance to prove them since my body condition is not fit enough to think that much.
1. The quadratic equation whose roots are k less than the roots of $$ax^2+bx+c=0$$ is $$a(x+k)^2+b(x+k)+c=0$$.
2. The quadratic equation whose roots are k more than the roots of $$ax^2+bx+c=0$$ is $$a(x-k)^2+b(x-k)+c=0$$.
3. The quadratic equation whose roots are n times the roots of $$ax^2+bx+c=0$$ is $$ax+bnx+cn^2=0$$.
4. The quadratic equation whose roots are negations of the roots of $$ax^2+bx+c=0$$ is $$ax^2-bx+c=0$$.
5. The quadratic equation whose roots are inverses of the roots of $$ax^2+bx+c=0$$ is $$cx^2+bx+a=0$$.
6. The quadratic equation whose roots are squareroots of the roots of $$ax^2+bx+c=0$$ is $$a^2x^2-(b^2-2ac)+c^2=0$$.
Thanks for your help.

 

[topsquark]

So, I found these statements and I need your assistance to prove them since my body condition is not fit enough to think that much.
1. The quadratic equation whose roots are k less than the roots of $$ax^2+bx+c=0$$ is $$a(x+k)^2+b(x+k)+c=0$$.
2. The quadratic equation whose roots are k more than the roots of $$ax^2+bx+c=0$$ is $$a(x-k)^2+b(x-k)+c=0$$.
3. The quadratic equation whose roots are n times the roots of $$ax^2+bx+c=0$$ is $$ax+bnx+cn^2=0$$.
4. The quadratic equation whose roots are negations of the roots of $$ax^2+bx+c=0$$ is $$ax^2-bx+c=0$$.
5. The quadratic equation whose roots are inverses of the roots of $$ax^2+bx+c=0$$ is $$cx^2+bx+a=0$$.
6. The quadratic equation whose roots are squareroots of the roots of $$ax^2+bx+c=0$$ is $$a^2x^2-(b^2-2ac)+c^2=0$$.
Thanks for your help.

Have you solved the quadratics for these to see? I'm using the "brute force" approach here. (If you would like to be more sophisticated you can use the Vieta formulas, but I don't think you gain any advantage using them.) If you like you can use the concept of translations and dilations of the graphs but I feel that working with the quadratic formula is the best bet for understanding the "why" of it. Your method of choice will be what unit you are currently studying.

1)
[math]ax^2 + bx + c = 0[/math] has roots [math]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math]

[math]a(x + k)^2 + b(x + k) + c = 0 \implies ax^2 + (2ak + b)x + (k^2 + bk + c) = 0[/math]

has roots [math]x = \dfrac{-(2ak + b) \pm \sqrt{ (2ak + b)^2 - 4a(k^2 + bk + c) }}{2a}[/math]

This simplifies to [math]x = \dfrac{-(2ak + b) \pm \sqrt{b^2 - 4ac}}{2a}[/math]

So is the conjecture true?

The rest are of a similar nature.

-Dan

 

[Monoxdifly]

I was stuck at exactly the line you ended in. And... I still don'e get the next line...

 

[topsquark]

I was stuck at exactly the line you ended in. And... I still don'e get the next line...

What is the difference between the roots?
[math]\dfrac{-(2ak + b) + \sqrt{b^2 - 4ac}}{2a} - \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} = [/math]?

-Dan

 

[HOI]

I don't see any reason to appeal to the quadratic formula. For the first one, suppose that u and v are roots to the equation $a(x+ k)^2+ b(x+ k)+ c= 0$. Then p= u+ k and q= v+ k satisfy $ax^2+ bx+ c= 0$.

 

[Olinguito]

In general:

If $p(x)=0$ is a polynomial equation of degree $n$ with roots $\alpha_1,\ldots,\alpha_n$, and $f:\mathbb R\to\mathbb R$ is an invertible function, then the polynomial equation whose roots are $f(\alpha_1),\ldots,f(\alpha_n)$ is
$$p\left(f^{-1}(x)\right)\ =\ 0.$$
Proof:

For each $i=1,\ldots,n$,
$$p\left(f^{-1}(f(\alpha_i))\right)\ =\ p(\alpha_i)\ =\ 0$$
as $\alpha_i$ is a root of $p(x)=0$.

Thus $\alpha$ is a root of $p(x)=0$ if and only if $f(\alpha)$ is a root of $p\circ f^{-1}(x)=0$.

This lemma applies to statements 1–5, where $p(x)=ax^2+bx+c$ and

1. $f(x)=x-k$,
2. $f(x)=x+k$,
3. $f(x)=nx$,
4. $f(x)=-x$,
5. $f(x)=\dfrac1x$.

Statement 6 is incorrect: it should read

6. The quadratic equation whose roots are squares of the roots of $$ax^2+bx+c=0$$ is $$a^2x^2-(b^2-2ac){\color{red}x}+c^2=0$$.

 

[Monoxdifly]

What is the difference between the roots?
[math]\dfrac{-(2ak + b) + \sqrt{b^2 - 4ac}}{2a} - \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} = [/math]?

–k?

I don't see any reason to appeal to the quadratic formula. For the first one, suppose that u and v are roots to the equation $a(x+ k)^2+ b(x+ k)+ c= 0$. Then p= u+ k and q= v+ k satisfy $ax^2+ bx+ c= 0$.

Gotta take notes to try using that someday.

In general:

If $p(x)=0$ is a polynomial equation of degree $n$ with roots $\alpha_1,\ldots,\alpha_n$, and $f:\mathbb R\to\mathbb R$ is an invertible function, then the polynomial equation whose roots are $f(\alpha_1),\ldots,f(\alpha_n)$ is
$$p\left(f^{-1}(x)\right)\ =\ 0.$$
Proof:

For each $i=1,\ldots,n$,
$$p\left(f^{-1}(f(\alpha_i))\right)\ =\ p(\alpha_i)\ =\ 0$$
as $\alpha_i$ is a root of $p(x)=0$.

Thus $\alpha$ is a root of $p(x)=0$ if and only if $f(\alpha)$ is a root of $p\circ f^{-1}(x)=0$.

This lemma applies to statements 1–5, where $p(x)=ax^2+bx+c$ and

1. $f(x)=x-k$,
2. $f(x)=x+k$,
3. $f(x)=nx$,
4. $f(x)=-x$,
5. $f(x)=\dfrac1x$.

Statement 6 is incorrect: it should read

Thank you. I did mistook "whose roots are squares" as "squareroots".