Question

if the root of x2+4mx+4m2+m+1=0 are real then

Answer 1

Answer:

The roots (zeros) are the

x

values where the graph intersects the x-axis. To find the roots (zeros), replace

y

with

0

and solve for

x

.

The roots cannot be found for this equation.

Step-by-step explanation:

Answer 2

The root of x2+4mx+4m2+m+1=0 is less than or equal to -1.

Given:

A quadratic equation x2+4mx+4m2+m+1 = 0 and its roots are real.

To Find:

The root of x2+4mx+4m2+m+1=0.

Solution:

To find the root of x2+4mx+4m2+m+1=0 we will follow the following steps:

As we know,

The above equation is quadratic because the maximum power of the variable (x) is 2 in this equation.

Now,

For the real roots,

Discriminant =

${b}^{2} - 4ac \geqslant 0$

a, b, are coefficients of x², x and c is the constant.

Now,

b = 4m

a = 1

c = 4m² + m + 1

Now,

In putting values we get,

${(4m)}^{2} - 4 \times 1 \times (4m² + m + 1) \geqslant 0$

$16{m}^{2} - 16 {m}^{2} - 4m - 4 \geqslant 0$

$- 4m - 4 \geqslant 0$

$- 4(m + 1) \geqslant 0$

$4(m + 1) \leqslant 0$

$m + 1 \leqslant 0$

$m \leqslant - 1$

Henceforth, The root of x2+4mx+4m2+m+1=0 is less than or equal to -1.

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