Question

given m is a real number not less than -1 such that this equation,
x²+2(m-2)x+m²-3m+3=0 has roots x₁,x₂ and x₁²+x₂² = 6
find value of m
pls answer it quickly

Answer 1

Answer:

(5 ± √17)/4

Step-by-step explanation:

Given Equation is x² + 2(m - 2)x + m² - 3m + 3 = 0.

On comparing with ax² + bx + c = 0, we get

a = 1, b = 2(m - 2), c = m² - 3m + 3.

Now,

(i) Sum of roots:

x₁ + x₂ = -b/a

x₁ + x₂ = -2(m - 2)

x₁ + x₂ = -2m + 4.

(ii) Product of roots:

x₁ * x₂ = c/a

x₁ * x₂ = (m² - 3m + 3)

On squaring (i) on both sides, we get

⇒ (x₁ + x₂)² = (-2m + 4)²

⇒ x₁² + x₂² + 2x₁x₂ = 4m² - 16m + 16

⇒ 6 + 2(m² - 3m + 3) = 4m² - 16m + 16

⇒ 6 + 2m² - 6m + 6 = 4m² - 16m + 16

⇒ 2m² + 10m - 4 = 4m²

⇒ -2m² + 10m - 4 = 0

⇒ 2m² - 10m + 4 = 0

Here, a = 2, b = -10, c = 4.

Now,

D = b² - 4ac

   = (-10)² - 4(2)(4)

   = 100 - 32

   = 68

The solutions are:

∴ x = -b ± √D/2a

     = (10) ± √68/(4)

     = (10 ± 2√17)/4

     = 5 ± √17/4.

Hope it helps!