## FIND THE QUADRATIC EQUATIONS WHOSE ROOTS ARE AND BETA

Question 1 :

If α and β are the roots of

2x2-3x-5  =  0

form a quadratic equation whose roots are α2 and β2

Solution :

By comparing the given equation with general form of quadratic equation we get,

a  =  2, b  =  -3 and c  =  -5

 Sum of the roots :α+β  =  -b/aα+β  =  3/2 Product of roots :α β  =  c/aαβ  =  -5/2

here α  =  α and β  =  β2

General form of quadratic equation whose roots are α2 and  β2

x2-(α22)x+α2β2  =  0 ---(1)

x2-(α22) x+(αβ)2  =  0

α22  =  (α+β)2-2αβ

=  (3/2)2 - 2 (-5/2)

=  (9/4) + 5

α22  =  29/4

By applying the values in (1), we get

x2-(29/4)x+(-5/2)2  =  0

4x2-29x+25  =  0

Therefore the required quadratic equation is

4x2-29x+25  =  0

Question 2 :

If α and β are the roots of

x2-3x+2  =  0

form a  quadratic equation whose roots are -α and -β.

Solution :

By comparing the given equation with general form of quadratic equation we get,

a  =  1, b  =  -3 and c  =  2

 Sum of the roots :α+β  =  -b/aα+β  =  3 Product of roots :αβ  =  c/aαβ  =  2

Here α  =  -α  and β  =  -β

General form of quadratic equation whose roots are α2 and β2

x2-(-α-β)x+(-α)(-β)  =  0

x2+(α+β)x+αβ  =  0  -----(1)

By applying the appropriate values in (1), we get

x2-3x+2  =  0

Therefore the required quadratic equations is

x2-3x+2  =  0

Question 3 :

If α and β are the roots of

x2-3x-1  =  0

then form a quadratic equation whose roots are 1/α2 and 1/β2.

Solution :

x2-3x-1  =  0

a = 1, b = -3 and c = -1

 Sum of the roots :α+β  =  -b/a= 3 Product of roots :αβ  =  c/a=  -1

Here α  =  1/α2and β  =  1/β2

By applying the above roots in the general form of a quadratic equation, we get

x2-(1/α2+1/β2)x+(1/α2) (1/β2)  =  0

x2-(α222β2)x+[(1/α)(1/β)]2  =  0

x2-[(α22)/(αβ)2]x+(1/αβ)2 = 0 ----(1)

α22  =  (α+β)2-2αβ

=  32-2(-1)

=  9+2

α22  =  11

By applying the value of α22 and αβ in (1), we get

x2-11x+1  =  0

So, the required quadratic equation is x2-11x+1  =  0.

Question 4 :

If α and β are the roots of the equation

3x2-6x+1  =  0

form an equation whose roots are

(i) 1/α , 1/β   (ii) α2β , β2α   (iii) 2α+β, 2β+α

Solution :

a  =  3, b  =  -6 and c  =  1

 Sum of the roots : α+β  =  -b/a ==>  6/3α+β  =  2 Product of roots αβ  =  c/aαβ  =  1/3

Here α = 1/α and β = 1/β

x2 - (1/α+1/β)x+(1/α) (1/β)  =  0

x2-[(α + β)/αβ]x+[(1/αβ)]  =  0  ----(1)

By applying the appropriate values in (1), we get

x2-6x+(2/3)  =  0

3x2-18x+2  =  0

So, the required quadratic equation is

3x2-18x+2  =  0

(ii) α2β , β2α

Solution :

Here α  =  α2β and β  =  β2α

Applying the given roots instead of α and β in the general form.

x2-(α2β+β2α)x+(α2β)(β2α)  =  0

x2-αβ(α+β)x+(α3β3)  =  0

x2-αβ(α+β)x+(αβ)3  =  0  ----(1)

By applying the appropriate values in (1), we get

x2-(2x/3)+(1/27)  =  0

27x2-18x+1  =  0

So, the required quadratic equation is

27x2-18x+1  =  0

(iii)  2α+β, 2β+α

Solution :

Here α  =  2α+β and β  =  2β+α

 α+β  =  2α+β+2β+α=  3α+3βα+β  =  3(α+β) αβ  =  (2α+β)(2β+α)  =  4αβ+2α²+2β²+αβαβ  =  5αβ+2(α2+β2)

x2-3(α+β)x+5αβ+2(α22)  =  0 ----(1)

α22  =  (α+β)2-2αβ

=  22 - 2 (1/3)

=  4-2/3

α22  =  10/3

By applying the values of α22α+β and αβ in (1), we get

x2-6x+(5/3)+(20/3) = 0

x2-6x+(25/3)  =  0

3x2-18x+25  =  0

So, the required quadratic equation is

3x2-18x+25  =  0 Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Kindly mail your feedback to v4formath@gmail.com

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