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

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 :

Roots of equation 2x² + 3x + 7 = 0 are α and β. The value of αβ^-1 + βα^-1 is

a)  2    b)  3/7     c)  7/2    d) -19/14

Solution :

2x² + 3x + 7 = 0

α/β + β/α = (α² + β²) / αβ

= [(α + β)² - 2 α β] / αβ  -----(1)

From this, we know that we have to find the sum and product of roots.

Sum of roots α + β = -b/a

Product of roots α β = c/a

a = 2, b = 3 and c = 7

α + β = -3/2 and α β = 7/2

Applying the above values in (1), we get

= [(-3/2)² - 2 (7/2)] / (7/2)

= [(9/4) - 7] / (7/2)

= [(9 - 28)/4] / (7/2)

= (-19/4) / (7/2)

= -19/4 x (2/7)

= -19/14

So, the value of αβ^-1 + βα^-1 -19/14.

Question 6 :

If α and β are the roots of the equation x² + 7x + 12 = 0 then the equation whose roots are (α + β)² and (α - β)² will be 

a)  x² - 14x + 49 = 0    b)  x² - 24x + 144 = 0

     c)  x² - 50x + 49 = 0   d) x² - 19x + 144 = 0

Solution :

x² + 7x + 12 = 0

Solving the given equation, we get

x² + 3x + 4x + 12 = 0

x(x + 3) + 4(x + 3) = 0

(x + 4) (x + 3) = 0

x = -4 and x = -3

α = -4 and β = -3

Values of (α + β)² and (α - β)² :

The required equation will be in the form,

x² - (sum of roots) x + Product of roots = 0

x² - (49 + 1) x + 49(1) = 0

x² - 50 x + 49 = 0

So, option c is correct.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.