Question

tell answer for 50 points ​

Answer 1

Given :-

If α and β are roots of the equation x² - 3x + 2 = 0

To Find :-

Equation whose roots are α + 1 and β + 1

Solution :-

⇒ x² - 3x + 2 = 0

⇒ x² - (x + 2x) + 2 = 0

⇒ x² - x - 2x + 2 = 0

⇒ x(x - 1) - 2(x - 1) = 0

⇒ (x - 1)(x - 2) = 0

⇒ x = 2 & 1

Now

⇒ α + 1

⇒ 2 + 1

⇒ 3

⇒ β + 1

⇒ 1 + 1

⇒ 2

Sum of zeroes :

⇒ 2 + 3

⇒ 5

Product of zeroes :

⇒ 2 × 3

⇒ 6

Now, equation is :

⇒ x² - (α + β)x + αβ

⇒ x² - 5x + 6

Answer 2

Answer:

α+β=3;αβ=2

S=(α+1+β+1)=(α+β+2)=5

P=(α+1)(β+1)=α+β+αβ+1

=3+2+1=6

∴ Equation whose roots are (α+1),(β+1) is x

2

−Sx+P=0 i.e., x

2

−5x+6=0