[ASK] Proof of Some Quadratic Functions

In summary: That's my mistake.So in summary, the quadratic equation can be manipulated in various ways to produce equations with roots that are translated, dilated, negated, or inverted versions of the original roots. This can be done by applying specific functions to the original equation, such as adding or subtracting constants, multiplying by a constant, or taking the reciprocal. However, statement 6 is incorrect and should read "The quadratic equation whose roots are squared versions of the roots of ax^2+bx+c=0 is a^2x^2-(b^2-2ac)x+c^2=0."
  • #1
Monoxdifly
MHB
284
0
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 \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle a(x+k)^2+b(x+k)+c=0\).
2. The quadratic equation whose roots are k more than the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle a(x-k)^2+b(x-k)+c=0\).
3. The quadratic equation whose roots are n times the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle ax+bnx+cn^2=0\).
4. The quadratic equation whose roots are negations of the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle ax^2-bx+c=0\).
5. The quadratic equation whose roots are inverses of the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle cx^2+bx+a=0\).
6. The quadratic equation whose roots are squareroots of the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle a^2x^2-(b^2-2ac)+c^2=0\).
Thanks for your help.
 
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  • #2
Monoxdifly said:
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 \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle a(x+k)^2+b(x+k)+c=0\).
2. The quadratic equation whose roots are k more than the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle a(x-k)^2+b(x-k)+c=0\).
3. The quadratic equation whose roots are n times the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle ax+bnx+cn^2=0\).
4. The quadratic equation whose roots are negations of the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle ax^2-bx+c=0\).
5. The quadratic equation whose roots are inverses of the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle cx^2+bx+a=0\).
6. The quadratic equation whose roots are squareroots of the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle a^2x^2-(b^2-2ac)+c^2=0\).
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)
\(\displaystyle ax^2 + bx + c = 0\) has roots \(\displaystyle x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(\displaystyle a(x + k)^2 + b(x + k) + c = 0 \implies ax^2 + (2ak + b)x + (k^2 + bk + c) = 0\)

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

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

So is the conjecture true?

The rest are of a similar nature.

-Dan
 
  • #3
I was stuck at exactly the line you ended in. And... I still don'e get the next line...
 
  • #4
Monoxdifly said:
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?
\(\displaystyle \dfrac{-(2ak + b) + \sqrt{b^2 - 4ac}}{2a} - \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} = \)?

-Dan
 
  • #5
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 $ax^2+ bx+ c= 0$.
 
  • #6
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

Monoxdifly said:
6. The quadratic equation whose roots are squares of the roots of \(\displaystyle ax^2+bx+c=0\) is \(\displaystyle a^2x^2-(b^2-2ac){\color{red}x}+c^2=0\).
 
  • #7
topsquark said:
What is the difference between the roots?
\(\displaystyle \dfrac{-(2ak + b) + \sqrt{b^2 - 4ac}}{2a} - \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} = \)?
–k?

Country Boy said:
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.

Olinguito said:
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".
 

1. What is a quadratic function?

A quadratic function is a polynomial function of degree 2, which means that the highest power of the variable is 2. It can be written in the form f(x) = ax^2 + bx + c, where a, b, and c are constants and x is the variable. Quadratic functions are often used to model relationships between two variables in mathematics and science.

2. How do I determine the vertex of a quadratic function?

The vertex of a quadratic function is the point where the function reaches its maximum or minimum value. To find the vertex, you can use the formula x = -b/2a, where a and b are the coefficients of the quadratic function. The y-coordinate of the vertex can be found by substituting the x-coordinate into the function.

3. What is the discriminant of a quadratic function?

The discriminant of a quadratic function is a value that can be calculated using the formula b^2 - 4ac. It is used to determine the number and nature of the solutions of a quadratic equation. If the discriminant is positive, the quadratic equation has two distinct real solutions. If it is zero, the equation has one real solution. And if it is negative, the equation has two complex solutions.

4. How do I graph a quadratic function?

To graph a quadratic function, you can plot a few points on a coordinate plane and connect them with a smooth curve. You can also use the vertex and the x-intercepts of the function to draw the graph. Remember to label the axes and include a scale to accurately represent the function.

5. What is the relationship between quadratic functions and parabolas?

A quadratic function is a mathematical representation of a parabola, which is a U-shaped curve. The graph of a quadratic function is always a parabola, and the properties of the function, such as the vertex and the x-intercepts, can be used to determine the properties of the parabola. Additionally, the equation of a parabola can be written in the form of a quadratic function.

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