FORMING NEW QUADRATIC EQUATIONS WITH RELATED ROOTS

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

here α  =  α²  and β  =  β²

General form of quadratic equation whose roots are α² and  β²

x²-(α²+β²)x+α²β²  =  0 ---(1)

x²-(α²+β²) x+(αβ)²  =  0

α²+β²  =  (α+β)²-2αβ

=  (3/2)² - 2 (-5/2)

=  (9/4) + 5

α²+β²  =  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

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

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-(α²+β²/α²β²)x+[(1/α)(1/β)]²  =  0

x²-[(α²+β²)/(αβ)²]x+(1/αβ)² = 0 ----(1)

α²+β²  =  (α+β)²-2αβ

=  3²-2(-1)

=  9+2

α²+β²  =  11

By applying the value of α²+β² 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 :

From the given quadratic equation

a  =  3, b  =  -6 and c  =  1

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

x²-3(α+β)x+5αβ+2(α²+β²)  =  0 ----(1)

α²+β²  =  (α+β)²-2αβ

=  2² - 2 (1/3)

=  4-2/3

α²+β²  =  10/3

By applying the values of α²+β², α+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

Question 5 :

Given that α + β = 5 and α β = -2, find the values of 

a)  (α / β) + (β / α)

b)  α³ + β³

c)  1/ α² + (1/ β²)

Solution :

a)  (α / β) + (β / α) = (α² + β²) / αβ

= [(α + β)² - 2 αβ] / αβ

= [(5)² - 2 (-2)] / (-2)

= (25 + 4) / (-2)

= -29/2

b)  α³ + β³ = [(α + β)³ -  3αβ((α + β)]

= 5³ -  3(-2)(5)

= 125 + 30

= 155

c)  1/ α² + (1/ β²) = (β² + α²) / α² β²

= (α²+ β²) / (α β)²

= [(α + β)² - 2 αβ]/ (α β)²

= [(5)² - 2 (-2)] / (-2)²

= (25+4) / 4

= 29/4

Question 6 :

Given that α + β = 5 and α β = 2/3, find the value of 

(α - β)²

Solution :

(α - β)² = (α + β)² - 2 αβ

= 5² - 2(2/3)

= 25 - (4/3)

= (75 - 4) / 3

= 71/3

So, the value of (α - β)² is 71/3.

Question 7:

Given that α + β = -3 and α β = 9, find the value of 

i)  α³β + β³α

ii) α / β  + β / α

Solution :

α + β = -3 and α β = 9

i)  α³β + β³α

= αβ(α² + β²)

= αβ[(α + β)² - 2αβ]

= 9[(-3)² - 2(9)]

= 9[9 - 18]

= 9(-9)

= -81

ii) α / β  + β / α = (α² + β²)/(α β)

From (i), we know that the value of α² + β² is -9

= -9/9

= -1

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