Question

Determine the value(s) of m for which the equation x2+ m(4x + m -1) + 2 =0 has real roots.
BEst answer will be brainliest
Q. no 11

Answer 1

When b^2-4ac = or > 0 then the equation has real roots
Here, a = 1 (coefficient of x^2)
b = 4m (coefficient of x )
c = 2 + m^2 - m ( constant term )

Filling the values
(4m)^2 - ( 4 × 1 × (2 + m^2 - m)) = or > 0
16m^2 - (8 + 4m^2 - 4m) = or > 0
12m^2 + 4m - 8 = or > 0

m = -b +-√b^2 - 4ac/2a
m = -4 +-√16 + 384/2×12


m = -4 + -√400/24
m = -4 + -20/24

With positive sign
m = -4 + 20/24
m = 16/24
m = 2/3

With negative sign
m = -4-20/24
m = -24/24
m = -1

So , m must be equal to or greater than -1 and 2/3

Answer 2

hello friend your answer is in the attachment