Question

If m,n are the roots of the equation ax^2+bx+c then find the roots of the equation cx^2+bx+a?

Answer 1

Answer:-`" "color(red)(1/n)` and `color(red)(1/m`

Explanation:

• If the roots of an equation `color(red)(ax^2+bx+c=0` is `color(blue)(alpha,beta`, then we can write as per rule that

  • `color(red)(alpha+beta)=-b/a`
  • `color(red)(alpha cdot beta)=c/a`

• As per given condition, we can write that

  • `color(red)(m+n)=-b/a`
  • `color(red)(m cdot n)=c/a`

• We will determine some values now for further use.

  • `color(red)(-b/c)=(-b/a)/(c/a)=(m+n)/(m cdot n)`
  • `color(red)(a/c)=1/(m cdot n`

• If the roots of the equation `color(red)(cx^2+bx+a=0` is `color(blue)(alpha,beta`, then,

  • `color(red)(alpha+beta)=-b/c=(m+n)/(m cdot n)" "...(1)`
  • `color(red)(alpha cdot beta)=a/c=1/(m cdot n`

• `color(red)(alpha-beta)`

`=sqrt((alpha+beta)^2-4 cdot alpha cdot beta)`

`=sqrt(((m+n)/(m cdot n))^2-4 /(m cdot n))`

`=(m-n)/(m cdot n)" "...(2)`

• From `(1)" & " (2)`,

  • `color(red)(alpha)=((m+n)/(m cdot n)+(m-n)/(m cdot n))/2=1/n`
  • `color(red)(beta)=((m+n)/(m cdot n)-(m-n)/(m cdot n))/2=1/m`

Hope this helps....
Thank you...

Answer 2

`1/m, 1/n`

Explanation:

Here's another in my series of pithy alternatives to overlong featured answers.

We're given `ax^2+bx+c` has zeros `m,n` and we're asked for the zeros of `cx^2+bx+a`.

It's pretty much implied `a ne 0` because we're given two roots. First let's do the case `c ne 0` so we're always dealing with real quadratic polynomials. That means `0` isn't a zero of either, i.e. `m ne 0, n ne 0.`

`am^2 + bm+ c = 0`

Dividing by `m^2` which we know isn't zero.

`1/m^2 (am^2 + bm + c) = 0`

`a + b/m + c/m^2 = 0`

`c(1/m)^2 + b(1/m) + a = 0`

So, `1/m` is a zero of `cx^2 + bx + a` as is of course `1/n.`

If `c=0` then one of `m` or `n` is zero, and the only the reciprocal of the other is the root of the linear equation `bx+a=0`.