Question

plz answer this question ​

Answer 1

Answer:

Step-by-step explanation:

in eq; $x^{2} + 4x - 3 = 0$

          $\alpha + \beta = -4\\\alpha.\beta = -3$

New eq has zeroes; $\frac{\alpha}{\beta }$ and $\frac{\alpha }{\beta }$

α/β + β/α = α²+β²/ αβ

α/β * β/α = αβ/αβ = 1

(a² + b²) = (a+b)²-2ab

wkt; α+β = -4  .... squaring both sides

       α²+β² = 10  

∴ α²+β²/ αβ = 10/-3

New eq :  K[x² - (sum of zeroes)x + (product of zeroes)]

              = x² - (-10/3)x + 1          .... taking LCM

              = (3x² + 10x + 3)/3    

              = 1/3 ( 3x² + 10x + 3)

Answer 2

Answer:

Step-by-step explanation:

For the equation X² + 4X - 3 = 0, the roots are ∝/β and β/∝

a=1, b=4, c= -3

∝+β = -b/a = -4/1 = -4 and ∝×β=∝β= c/a = -3/1 = -3

(∝+β)² = (∝+β)(∝+β)

(∝+β)² = ∝² + ∝β + ∝β + β²

(∝+β)² = ∝² + β² + 2∝β

(∝+β)² - 2∝β = ∝²+β²

The equation formula is : X² - (sum of roots)X + (product of roots) = 0

sum of roots = (∝/β) + (β/∝)

= (∝²+β²)/∝β

= [(∝+β)² - 2∝β) ]/∝β

= [(-4)² - 2(-3) ]/(-3)

= (16 + 6)/ -3

= 22/ -3

product of roots = (∝/β) × (β/∝)

= ∝β/∝β

= 1

∴ the equation = X² - (-22/3)X + 1 = 0

⇒ X² + (11/3)X + 1 = 0

⇒ 3X² + 22X + 3 = 0