The page relations between roots solution6 is continuation of page5 containing solution of some practice questions from the worksheet relationship between roots and coefficients.

(ii) α² β , β² α

**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 = α² β and β = β² α

General form of quadratic equation whose roots are α² β and β² α

x² - (α² β + β² α) x + (α² β) (β² α) = 0

x² - α β (α + β) x + (α³ β³)] = 0

x² - α β (α + β) x + (α β)³ = 0

x² - (1/3) (2) x + (1/3)³ = 0

x² - (2 x/3) + (1/27) = 0

(27 x² - 18 x + 1)/27 = 0

27 x² - 18 x + 1 = 0

Therefore the required quadratic equation is 27 x² - 18 x + 1 = 0

(iii) 2 α + β , 2 β + α

here α = 2 α + β and β = 2 β + α

General form of quadratic equation whose roots are 2 α + β and 2 β + α

x² - (2 α + β + 2 β + α) x + (2 α + β) (2 β + α) = 0

x² - (3 α + 3 β) x + (4 α β + 2 α² + 2 β² + α β) = 0

x² - 3 (α + β) x + [5 α β + 2(α² + β²) ] = 0

α² + β² = (α + β)² - 2 α β

= (2)² - 2 (1/3)

= 4 - 2/3

= (12 - 2)/3

= 10/3

x² - 3 (2) x + [5 (1/3) + 2(10/3)] = 0

x² - 6 x + [(5/3) + 20/3] = 0

x² - 6 x + [25/3] = 0

(3 x² - 18 x + 25)/3 = 0

3 x² - 18 x + 25 = 0

Therefore the required quadratic equation is 3 x² - 18 x + 25 = 0.

These are the problems solved in the page relations between roots solution6.

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

relations between roots solution6

- Practice problems on nature of roots
- Practical problems in quadratic equation
- Framing quadratic equation from roots
- Square root
- Solving linear equation in cross multiple method
- Solving linear equations in elimination method