Question

Determine the value(s) of m for which the equation
x^2 + m(4x + m - 1) + 2 =0 has real roots.​

Answer 1

Answer: m ≥ 2/3 or m ≥ -1

Step-by-step explanation:

x^2 + m(4x + m - 1) + 2 =0

x^2 + 4mx + m^2- m + 2 =0

Here, a= 1, b = 4m, c = m^2- m + 2,

For real roots, we have D ≥ 0,

D= b^2-4ac ≥ 0

(4m)^2-4×1× (m^2- m + 2) ≥ 0

16m^2 -4m^2 + 4m -8 ≥ 0

12m^2 +4m - 8 ≥ 0

Divide both sides by 4,

3m^2+ m - 2 ≥ 0

3m^2+ 3m-2m - 2 ≥ 0

3m(m+1) -2(m+1) ≥ 0

(3m-2) (m+1) ≥ 0

Either 3m-2 ≥ 0, m ≥ 2/3

Or, m+1 ≥ 0

m ≥ -1