I’ve watched R2drew yt video but fro question 1, yotta says in the comments that u can solve it by looking at the coefficients of x^5 and x^6. Not sure how that works. Help pls

Original post by Fndjdidisjb

I’ve watched R2drew yt video but fro question 1, yotta says in the comments that u can solve it by looking at the coefficients of x^5 and x^6. Not sure how that works. Help pls

If you have a quadratic like

x^2 + 3x + 2 = 0

roots are -2 and -1 and 3 is their negaitve sum. So if you have a

(x-r1)(x-r2)(x-r3)(x-r4)(x-r5)(x-r6) = 0

what is the coeficient of x^5 and what would you do if the x^6 term is not unity?

Original post by mqb2766

If you have a quadratic like

x^2 + 3x + 2 = 0

roots are -2 and -1 and 3 is their negaitve sum. So if you have a

(x-r1)(x-r2)(x-r3)(x-r4)(x-r5)(x-r6) = 0

what is the coeficient of x^5 and what would you do if the x^6 term is not unity?

x^2 + 3x + 2 = 0

roots are -2 and -1 and 3 is their negaitve sum. So if you have a

(x-r1)(x-r2)(x-r3)(x-r4)(x-r5)(x-r6) = 0

what is the coeficient of x^5 and what would you do if the x^6 term is not unity?

Coefficient of x^5 is -15 and idk what u mean by unity for x^6?

How does that help with finding points of intersection with the line?

(edited 4 months ago)

Original post by Fndjdidisjb

Coefficient of x^5 is -15 and idk what u mean by unity for x^6?

How does that help with finding points of intersection with the line?

How does that help with finding points of intersection with the line?

The intersection of the line does not affect the x^5 and x^6 terms, only the constant and the linear terms, so its basically the same as the basic root finding problem in terms of the sum of the roots. So the sum of the roots is not changed, even though the individual roots will be (as the constant and linear terms are altered). So youd get a 6th order polynomial like

pi x^6 - pi 15 x^5 + .... = 0

Dividing by pi to make the x^6 coefiicient unity means the x*5 coefficient is the (negated) sum of the points of intersection

Original post by mqb2766

The intersection of the line does not affect the x^5 and x^6 terms, only the constant and the linear terms, so its basically the same as the basic root finding problem in terms of the sum of the roots. So the sum of the roots is not changed, even though the individual roots will be (as the constant and linear terms are altered). So youd get a 6th order polynomial like

pi x^6 - pi 15 x^5 + .... = 0

Dividing by pi to make the x^6 coefiicient unity means the x*5 coefficient is the (negated) sum of the points of intersection

pi x^6 - pi 15 x^5 + .... = 0

Dividing by pi to make the x^6 coefiicient unity means the x*5 coefficient is the (negated) sum of the points of intersection

I understand now. Ty

Original post by Fndjdidisjb

I understand now. Ty

NP. Its worth having a simple example in your pocket so say you had a quadratic

y = ax^2 + bx + c

intersecting a horizontal line

y = d

The sum of the roots of the original quadratic is -b/a (twice the stationary point -b/2a), which you can think about algebraically as the quadratic is

a(x-r1)(x-r2)

or by noting that the quadratic formula gives

x = -b/2a +/-....

When you add the two roots, the ... cancels and r1+r2 = -b/a, so twice the stationary point or twice the average of the roots.

This holds for the sum of the intersection points as the quadratic is unchanged apart from the constant becomes c-d, so the average (sum) of the roots does not change its -b/2a. Its independent of the value of the constant c.

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