How to deducec(x−1α)(x−1β)fromcx2+bx+awhen we dont know if 1α and 1β are the roots of equation cx2+bx+a

Step 1By Vieta's formulasα+β=−ba and αβ=caThenc(x−1α)(x−1β)=c(x2−α+βαβx+1αβ)=c(x2+bcx+ac)=cx2+bx+aStep 2Here is the general situation:Let f(x)=anxn+an−1xn−1+⋯a1x+a0 with n≥1 and an≠0. Assume also that a0≠0. This is the same as assuming that 0 is not a root of f. Let α1,⋯,an be the roots of f (they are not necessarily distinct). Note thatf(x)=an(x−α1)⋯(x−αn)Let g(x)=f(1x). Note that the roots of g are 1α1,⋯,1αnSinceg(x)−f(1x)=1xn(a0xn+a1xn−1+…an−1x+an)the roots of the polynomiala0xn+a1xn−1+…an−1x+an are precisely 1α1,…,1αn and hencea0xn+a1xn−1+…an−1x+an=a0(x−1α1)…(x−1αn)

Step 1We havec1α2+b1α+a=1α2(c+bα+aα2)=0.Hence 1α2 is a root of the equation cx2+bx+a=0

Step 1By Vieta's formulasα+β=−ba and αβ=caThenc(x−1α)(x−1β)=c(x2−α+βαβx+1αβ)=c(x2+bcx+ac)=cx2+bx+aStep 2Here is the general situation:Let f(x)=anxn+an−1xn−1+⋯a1x+a0 with n≥1 and an≠0. Assume also that a0≠0. This is the same as assuming that 0 is not a root of f. Let α1,⋯,an be the roots of f (they are not necessarily distinct). Note thatf(x)=an(x−α1)⋯(x−αn)Let g(x)=f(1x). Note that the roots of g are 1α1,⋯,1αnSinceg(x)−f(1x)=1xn(a0xn+a1xn−1+…an−1x+an)the roots of the polynomiala0xn+a1xn−1+…an−1x+an are precisely 1α1,…,1αn and hencea0xn+a1xn−1+…an−1x+an=a0(x−1α1)…(x−1αn)

Step 1We havec1α2+b1α+a=1α2(c+bα+aα2)=0.Hence 1α2 is a root of the equation cx2+bx+a=0

Expert Answer to "How to deducec(x−1α)(x−1β)fromcx2+bx+awhen we dont know if 1α..."

High School
Algebra
Algebra I
Quadratic function and equation

NepanitaNesg3a

Answered question

2022-04-23

How to deduce

c (x - \frac{1}{α}) (x - \frac{1}{β})

from

c x^{2} + b x + a

when we dont know if

\frac{1}{α}

and

\frac{1}{β}

are the roots of equation

c x^{2} + b x + a

Answer & Explanation

Aliana Kaufman

Beginner2022-04-24Added 13 answers

Step 1
By Vieta's formulas

α + β = \frac{- b}{a}

and

α β = \frac{c}{a}

Then

c (x - \frac{1}{α}) (x - \frac{1}{β}) = c (x^{2} - \frac{α + β}{α β} x + \frac{1}{α β})

= c (x^{2} + \frac{b}{c} x + \frac{a}{c}) = c x^{2} + b x + a

Step 2
Here is the general situation:
Let

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots a_{1} x + a_{0}

with

n \geq 1

and

a_{n} \neq 0

. Assume also that

a_{0} \neq 0

. This is the same as assuming that 0 is not a root of f. Let

α_{1}, \dots, a_{n}

be the roots of f (they are not necessarily distinct). Note that

f (x) = a_{n} (x - α_{1}) \dots (x - α_{n})

Let

g (x) = f (\frac{1}{x})

. Note that the roots of g are

\frac{1}{α_{1}}, \dots, \frac{1}{α_{n}}

Since

g (x) - f (\frac{1}{x})

= \frac{1}{x^{n}} (a_{0} x^{n} + a_{1} x^{n - 1} + \dots a_{n - 1} x + a_{n})

the roots of the polynomial

a_{0} x^{n} + a_{1} x^{n - 1} + \dots a_{n - 1} x + a_{n}

are precisely

\frac{1}{α_{1}}, \dots, \frac{1}{α_{n}}

and hence

a_{0} x^{n} + a_{1} x^{n - 1} + \dots a_{n - 1} x + a_{n} = a_{0} (x - \frac{1}{α_{1}})

\dots (x - \frac{1}{α_{n}})

belamontern9i

Beginner2022-04-25Added 19 answers

Step 1
We have

c \frac{1}{α^{2}} + b \frac{1}{α} + a = \frac{1}{α^{2}} (c + b α + a α^{2}) = 0 .

Hence

\frac{1}{α_{2}}

is a root of the equation

c x^{2} + b x + a = 0

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Algebra I

Didn't find what you were looking for?