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if alpha and beta are the zeroes of a quadratic polynomial 2x^2 - 3x - 5 form a polynomial whose zeroes are (i)alpha^2 and beta^2 (ii)1/alpha^2 and 1/beta^2

Guest Apr 29, 2015

Best Answer 

 #1
avatar+26329 
+10

The polynomial can be factored into: (x+1)(2x-5), so we have α = -1 and β = 5/2

 

(i)  The polynomial with zeros α2 and β2 can be factored as (x - α2)(x - β2).   This can be multiplied out to get 

x2 - (α2 + β2)x + α2β2    or    x2 - (1 + 25/4)x + 25/4    or    x2 -  (29/4)x + 25/4  

 

Multiply all the terms by 4 to get   4x2 -  29x + 25

 

See if you can now do part (ii).

Alan  Apr 29, 2015
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2+0 Answers

 #1
avatar+26329 
+10
Best Answer

The polynomial can be factored into: (x+1)(2x-5), so we have α = -1 and β = 5/2

 

(i)  The polynomial with zeros α2 and β2 can be factored as (x - α2)(x - β2).   This can be multiplied out to get 

x2 - (α2 + β2)x + α2β2    or    x2 - (1 + 25/4)x + 25/4    or    x2 -  (29/4)x + 25/4  

 

Multiply all the terms by 4 to get   4x2 -  29x + 25

 

See if you can now do part (ii).

Alan  Apr 29, 2015
 #2
avatar+91049 
+5

 

If alpha and beta are the zeroes of a quadratic polynomial  $$2x^2 - 3x - 5$$  form a polynomial whose zeroes are

The roots for this is  2.5 an -1         These are    $$\alpha\;\;and\;\;\beta$$

 

(i)      $$\alpha^2=6.25\;\;and \;\;\beta^2 =1$$

 

In any quadratic    ax^2+bx+c   the sum of the roots is given by  -b/a = 7.25    

and the product of the roots is c/a = 6.25

So one solution would be       $$x^2-7.25x+6.25$$

 

 

 

(ii)     $$\frac{1}{\alpha^2} \;\;and\;\; \frac{1}{\beta^2}$$       

 

   $$\\Roots\;are\;\frac{1}{6.25}=\frac{4}{25}=0.16\;\;and\;\;1\\\\
so\;\;\frac{-b}{a}=1.16\;\;and\;\; \frac{c}{a}=0.16\\
Let a=1\\
$So one solution would be $ x^2-1.16x+0.16$$

Melody  Apr 30, 2015

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