Question

If equation x2 – (2 + m)x + (– m2 – 4m – 4) = 0 has coincident roots, then find the value of m.

Answer 1

Answer:

The value of m is -2, -2.

Step-by-step explanation:

The general form of the equation is $ax ^2+bx+c=0$, if the roots are equal, then $b ^2-4ac=0$.

Consider the equation

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

Here,

$a=1\\\\b=-(2+m)\\\\c=(-m ^2 -4m-4)$

Substituting the values of a, b, and c in  $b ^2-4ac=0$ we get,

$(2+m) ^2 -4 (-m ^2-4m-4) = 0\\\\4+m ^2+4m+4m ^2 +16m+16 = 0\\\\5m^2+20m+20 = 0\\\\m^2+4m+4 = 0\\\\(m+2)^2 = 0\\\\m = -2, -2.$

Answer 2

Answer:

m = 6, m = 2/3

Step-by-step explanation:

 The given quadratic equation is x² - (2+ m)x + (m² -4m + 4) = 0

comparing it with ax² + bx + c = 0

a = 1, b = -(2 + m), c = (m² - 4m + 4)

Since the equation has coincident roots

b² - 4ac = 0

[-(2 + m)]² - 4 X 1 X (m² - 4m + 4) = 0

4 + 4m + m² - 4m² + 16m - 16 = 0

- 3m² + 20m - 12 = 0

Or, 3m² - 20m + 12 = 0

3m² -18m - 2m + 12 = 0

3m(m - 6) - 2(m - 6) = 0

m- 6 = 0

m = 6

3m - 2 = 0

3m = 2

m = 2/3.