Question

if alpha +beta =3 and alpha3 +beta3=7, then alpha and beta are the roots of equation
a) 9x2 +27x+20=0,b) 9x ​2 -27x+20=0, c) 9x2+27x-20=0 and d) 9x2 -27x-20 =0

and in this either b or d will be the answer and i need the complete solution

Answer 1

Let, alpha = a and beta = b
Now, a + b = 3
a³ + b³ = 7
Solving identity, a³ + b³ = (a+b) (a² + b² - ab) =7
Here, a+b = 3
3(a² + b ² - ab) = 7
Now, a² + b² = (a+b)² - 2ab
Now, 3((a + b) ² - 2ab) = 7
3 ((3)² - 2ab) =7
3( 9 - 2ab) = 7
27 - 6ab = 7
- 6ab = 7 - 27
- 6ab = - 20
ab = 20/6 = 10/3
Now, a+b = 3 and ab = 10/3
Equation = k( x² - (a+b) x + ab)
= k( x² - 3x + 10/3)
= k ( (3x² - 9x + 10) / 3)
Put, k=3
Now, equation = 3x² - 9x + 10
Now, 3( 3x² - 9x + 10)
9x² - 27 x + 30

Answer 2

Answer:

Step-by-step explanation:a+b=3

a^3+b^3=7

Therefore a^3+b^3=(a+b)(a^2+b^2-ab) =7

Here a+b=3

Therefore 3(a^2+b^2-ab) =7

3((a+b)^2 - 2ab -ab) =7

3((3)^2- 3ab) =7

27-9ab=7

ab=20/9

Therefore x^2-(a+b)x+ab

x^2-3x+20/9

=9x^2-27x+20/9 = 0

=9x^2-27x+20=0