Question

If alpha, beta be the roots of the equation 3x^2+2x+1=0, find an equation whose roots are 1-alpha/1+alpha, 1-beta/1+beta.

Answer 1

Answer: acx2+b(a+c)x+(a2+c2+2ac)=0acx2+b(a+c)x+(a2+c2+2ac)=0 .

Solution:

The given quadratic equation is ax2+bx+c=0ax2+bx+c=0  with roots αα and ββ . Let, sum of roots = α+β=p=−baα+β=p=−ba  and product of roots = αβ=r=caαβ=r=ca

We have to find the equation whose roots are β+1αβ+1α and α+1β,α+1β,

or, the new roots should be αβ+1ααβ+1α  and  αβ+1βαβ+1β

or, the new roots should be r+1αr+1α  and  r+1βr+1β .

So, basically the new root is r+1xr+1x,  where r = product of roots = αβ=caαβ=caand x takes the values αα and ββ.

Let, the new root be represented by yy , so that  y=r+1xorx=r+1yy=r+1xorx=r+1y

Setting this value, x=r+1yx=r+1y into the equation ax2+bx+c=0,ax2+bx+c=0,  we get

a(r+1y)2+b(r+1y)+c=0a(r+1y)2+b(r+1y)+c=0

or,  cy2+b(r+1)y+a(r+1)2=0cy2+b(r+1)y+a(r+1)2=0

Putting r=car=ca,  we get

cy2+b(ca+1)y+a(ca+1)2=0,cy2+b(ca+1)y+a(ca+1)2=0, 

or, cy2+b(c+aa)y+a(c+aa)2y=0cy2+b(c+aa)y+a(c+aa)2y=0,

or, cy2+b(c+aa)y+(c+a)2a=0,cy2+b(c+aa)y+(c+a)2a=0,

or, acy2+b(a+c)y+(a2+c2+2ac)=0acy2+b(a+c)y+(a2+c2+2ac)=0.

Replacing yy with xx in order to write the equation in traditional way, we get the newly formed equation as the answer:

acx2+b(a+c)x+(a2+c2+2ac)=0acx2+b(a+c)x+(a2+c2+2ac)=0  Ans.

Answer 2

Answer:

Step-by-step explanation:

f(x) = 3x²+2x+1

roots are a and b

new zeroes 1-a/1+a   and 1-b/1+b

trick :  x = 1-a/1+a

          x + xa = 1-a

           x+ xa - 1 +a = 0

       a(x+1) = 1-x

      a = 1-x/1+x

now replace x by 1-x/1+x in f(x)

3 (1-x/1+x)² + 2(1-x/1+x) + 1 = 0

x² -2x +3 =0 is the required polynomial.

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Hope it helps!