## Relations Between Roots Solution5

The page relations between roots solution5 is containing solution of some practice questions from the worksheet relationship between roots and coefficients.

(7) If α and β are the roots of x² - 3 x - 1 = 0, then form a quadratic equation whose roots are 1/α² and 1/β².

Solution:

General form of quadratic equation whose roots are α and β

x² - (α + β) x + α β = 0

by comparing the given equation with general form of quadratic equation we get a = 1  b = -3 and c = -1

Sum of the roots α + β = -b/a

= -(-3)/1

= 3

Product of roots α β = c/a

= -1/1

= -1

here α = 1/α²  and β = 1/β²

General form of quadratic equation whose roots are α² and β²

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/α)(1/β)]² = 0

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

α² + β² = (α + β)² - 2 α β

= (3)² - 2 (-1)

= 9 + 2

= 11

x² - [11/(-1)²] x + [(1/-1) = 0

x² - 11 x + 1 = 0

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

(8) If α and β are the roots of the equation 3 x² - 6 x + 1 =0,form an equation whose roots are

(i) 1/α , 1/β

Solution:

General form of quadratic equation whose roots are α and β

x² - (α + β) x + α β = 0

by comparing the given equation with general form of quadratic equation we get a = 3  b = -6 and c = 1

Sum of the roots α + β = -b/a

= -(-6)/3

= 2

Product of roots α β = c/a

= 1/3

here α = 1/α and β = 1/β

General form of quadratic equation whose roots are 1/α and 1/β

x² - (1/α + 1/β) x + (1/α) (1/β) = 0

x² - [(α + β)/αβ] x + [(1/αβ)] = 0

x² - [2/(1/3)] x + [2 (1/3)] = 0

x² - [2 x (3/1)] x + [2/3] = 0

x² - 6 x + (2/3) = 0

3 x² - 18 x + 2 = 0

Therefore the required quadratic equation is 3 x² - 18 x + 2 = 0.

(ii) α² β , β² α

(iii) 2 α  + β , 2 β + α

These are the problems solved in the page relations between roots solution5.

You can find solution of other problems in the next page.