Consider equation ax^2+bx+c=0 whose roots are x and y.Then find quadratic equation whose roots are x/y and y/x?

Answer 1

# acx^2-(b^2-2ac)x+ac=0#.

Given that, #x and y# are the roots of the quadr. eqn. :
#ax^2+bx+c=0#.
#:. x+y=-b/a, and, xy=c/a....................(ast)#.
Let, #X=x/y, and, Y=y/x#.
Then, #X+Y=x/y+y/x=(x^2+y^2)/(xy)#,
#=1/(xy){(x+y)^2-2xy}#,
#=(x+y)^2/(xy)-2#,
#=((-b/a)^2)/(c/a)-2#.
#:. X+Y=(b^2-2ac)/(ac).................(ast^1)#.
Also, #X*Y=1..................................(ast^2)#.

Hence, the desired quadr. eqn. is given by,

#x^2-(X+Y)x+XY=0,#
# i.e., x^2-((b^2-2ac)/(ac))x+1=0,#
# or, acx^2-(b^2-2ac)x+ac=0#.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

The equation is #acx^2-(b^2-2ac)x+ac=0#

Let the roots of the equation

#ax^2+bx+c=0#

be

#alpha# and #beta#

Then,

#alpha+beta=-b/a#

and

#alphabeta=c/a#

The roots of the new equation are

#alpha/beta# and #beta/alpha#

Then the sum of the roots are

#alpha/beta+beta/alpha=(alpha^2+beta^2)/(alphabeta)#
#=((alpha+beta)^2- 2alphabeta)/(alphabeta)#
#=((-b/a)^2-2*c/a)/(c/a)#
#=(b^2/a^2-2c/a)/(c/a)#
#=(b^2-2ac)/(ac)#

The product of the roots is

#alpha/beta*beta/alpha=1#

The quadratic equation is

#x^2-((b^2-2ac)/(ac))x+1=0#
#acx^2-(b^2-2ac)x+ac=0#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

The quadratic equation whose roots are (x/y) and (y/x) can be obtained by constructing a new equation using Vieta's formulas, which state that for a quadratic equation (ax^2 + bx + c = 0), the sum of the roots is (-b/a) and the product of the roots is (c/a).

Given that the roots of the original equation (ax^2 + bx + c = 0) are (x) and (y), we have:

Sum of the roots: (x + y = -\frac{b}{a})

Product of the roots: (xy = \frac{c}{a})

From the given roots (x/y) and (y/x), we can express (x) and (y) in terms of each other:

(x = ky) and (y = \frac{x}{k}), where (k) is a constant.

Substituting these expressions into the equations for the sum and product of roots:

(x + y = ky + \frac{x}{k} = -\frac{b}{a})

(xy = \left(ky\right) \left(\frac{x}{k}\right) = \frac{c}{a})

Solving these equations simultaneously, we find (k):

(ky + \frac{x}{k} = ky + \frac{ky}{k} = ky + y = \frac{ky(b - a)}{a} = -\frac{b}{a})

(\Rightarrow ky(b - a) + ay = -b)

(\Rightarrow y(kb - ka + a) = -b)

(\Rightarrow y = -\frac{b}{kb - ka + a})

Since (y = \frac{x}{k}):

(\frac{x}{k} = -\frac{b}{kb - ka + a})

(\Rightarrow x = -\frac{kb}{kb - ka + a})

Now, we have expressions for (x) and (y). We can construct the new quadratic equation using these roots:

(ax^2 + bx + c = a \left(-\frac{kb}{kb - ka + a}\right)^2 + b \left(-\frac{kb}{kb - ka + a}\right) + c)

(= a\left(\frac{k^2b^2}{(kb - ka + a)^2}\right) - b\left(\frac{k^2b}{kb - ka + a}\right) + c)

(= \frac{ak^2b^2}{(kb - ka + a)^2} - \frac{bk^2b}{kb - ka + a} + c)

This equation represents the quadratic equation with roots (x/y) and (y/x).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7