Root of an equation 'x1,x2'

C

Captain Kirk

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Hello, I have a question. How does it call the equation that has a root of an equation 'x1,x2'? I am trying to find it everywhere but, nothing, which is weird. I would like to learn about it to understand it. Thank you for reading.

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If [math]x_1[/math] and [math]x_2[/math] are roots of a quadratic equation then [math](x - x_1)(x - x_2) = 0[/math]. Is that your question?

-Dan
 
topsquark said:
If [math]x_1[/math] and [math]x_2[/math] are roots of a quadratic equation then [math](x - x_1)(x - x_2) = 0[/math]. Is that your question?

-Dan
Click to expand...
If that's the formula, then yes. Do you know the formula's name? I dont find any topic about it
 
Dan

I think the question is about the roots of

[MATH]x^2(x_1 + x_2) - 2x(x_1 + x_2) + x_1x_2(x_1 + x_2)\\ \text {given that } x_1 \text { and } x_2 \text { are roots of } x = ax^2 + bx + c \text { and that } x_1x_2 < 1.[/MATH]The notation is a bit convoluted (but that may be part of the point).

If I am correct, then the first thing I would do is to replace x1 and x2 with p and q (because I would screw up those subscripts somewhere) and then state p and q in terms of a, b, and c.

I suspect everything simplifies out a great deal.
 
Actually, taking my own advice

[MATH]f(x) = p(x - q)^2 + q(x - p)^2 = p(x^2 - 2qx + q^2) + q(x^2 - 2px + p^2 =\\ (p + q)x^2 - 2x(pq + pq) + pq^2 + qp^2 = (p + q)x^2 + 4pqx + pq(p + q).[/MATH]
Plus in addition to the points about p and q being the roots of ax2 + bx + c and pq < 1, I think we are to assume that a, b, and c, are all real numbers.
 
Note [MATH]x_1 x_2<0[/MATH] where [MATH]x_1[/MATH], [MATH]x_2[/MATH] are roots of a quadratic, we assume with real coefficients.
What can you say about the roots of the this quadratic: [MATH]x_1(X-x_2)^2 + x_2(X-x_1)^2=0[/MATH]?

Multiply out the quadratic:
[MATH]x_1 X^2 -2x_1x_2X+x_1x_2^{2}+ x_2 X^2 -2x_2x_1X+x_2x_1^{2}=0\\ (x_1+x_2)X^2 -4x_1x_2X+x_1x_2^{2}+x_2x_1^{2}=0\\ (x_1+x_2)X^2 -4x_1x_2X+x_1x_2(x_1+x_2)=0 \hspace2ex \text{(*)}[/MATH]
If [MATH](x_1+x_2)≠0[/MATH] you can divide by [MATH](x_1+x_2)[/MATH] and get:

[MATH]X^2-\frac{4x_1x_2}{x_1+x_2}X+x_1x_2=0[/MATH]
Hopefully you know the relationship between the coefficients and the roots of a quadratic and can answer the question reasonably now.
(Vieta's Formula)
 
Glad you have it solved.
Here is the rest of the solution I started earlier.

Note [MATH]x_1 x_2<0[/MATH] where [MATH]x_1[/MATH], [MATH]x_2[/MATH] are roots of a quadratic, we assume with real coefficients.
What can you say about the roots of the this quadratic: [MATH]x_1(X-x_2)^2 + x_2(X-x_1)^2=0[/MATH]?

Multiply out the quadratic:
[MATH]x_1 X^2 -2x_1x_2X+x_1x_2^{2}+ x_2 X^2 -2x_2x_1X+x_2x_1^{2}=0\\ (x_1+x_2)X^2 -4x_1x_2X+x_1x_2^{2}+x_2x_1^{2}=0\\ (x_1+x_2)X^2 -4x_1x_2X+x_1x_2(x_1+x_2)=0 \hspace2ex \text{(*)}[/MATH]
Now, if [MATH] (x_1+x_2)=0[/MATH], i.e. [MATH]x_1=-x_2[/MATH], then (*) is a linear equation and has only one solution, X=0
(which is not given as one of your answer options)!

If [MATH](x_1+x_2)≠0[/MATH], then the quadratic equation (*) becomes:
[MATH]X^2-\frac{4x_1x_2}{x_1+x_2}X+x_1x_2=0[/MATH](Note this is a quadratic with real coefficients, since [MATH]x_1[/MATH] and [MATH]x_2[/MATH] are roots of a quadratic with real coefficients, therefore [MATH](x_1+x_2)[/MATH] is real and we know [MATH]x_1x_2<[/MATH]0).

So the product of the roots of this equation is [MATH]x_1x_2<0[/MATH] (the constant term in the quadratic).
Note the roots of this quadratic are real since [MATH]x_1x_2<0[/MATH] meaning the discriminant is positive.
i.e. the roots are real with opposite signs.

(Note: the answers omit the option where [MATH]x_1=-x_2≠0[/MATH] when there is one root, X=0).


(Vieta's Formula)
 
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