Question

If the roots of x square +4mx + 4m square +m+1=0 are real, which of the following is true ? A m=-1 B m= smaller than equal to -1 C m= greater than equal to -1 D m= greater than equal to 0

Answer 1

Answer:

m < -1

Step-by-step explanation:

Equation: $x^{2}$ + 4m$x$ + 4$m^{2}$ + m +  1 = 0

We Know That if equation has real roots,

Then Discriminant is greater that zero (D > 0)  ----------> 1

Discriminant = $(4m)^{2}$ - 4(1)(4$m^{2}$ + m +  1)   <---------------------- ($b^{2}$ - 4ac)

                     = 16$m^{2}$ - (16$m^{2}$ + 4m + 4)

                     = 16$m^{2}$ - 16$m^{2}$ - 4m - 4

                     = -4m - 4

From 1,

-4m - 4 > 0

- (4m + 4) > 0

4m + 4 < 0 (sign changes if we move -ve from one side to another)

4m < -4

m < -1 (no option given is correct)

I think by seeing the options, that the question would be if the roots are equal and real. In this case, m = -1 since D = 0. So A is answer, but when D > 0 , no option is correct.

HOPE IT HELPS!!!