Verified answer

Step-by-step explanation:

Given:

[tex]mx^{2} - 2(m-1)x + (m+2) = 0[/tex]

For a quadratic equation to have real and equal roots,

[tex]b^{2} - 4ac = 0[/tex]

So,

[tex](2(m-1))^{2} - 4 ( m+2)(m) = 0 \\\\(2m-2)^{2} - 4(m^{2} + 2m) = 0 \\ \\4m^{2} - 8m + 4 - 4m^{2} - 8m = 0 \\\\-8m = -4 \\\\m = \frac{4}{8} \\\\m = \frac{1}{2}[/tex]

Answer:

m = [tex]\frac{1}{2}[/tex]

Answer:

mx²-2(m-1)x+m+2=0

Step-by-step explanation:

Here,a=m,b=-2(m-1),c=(m+2).

D=b²-4ac

Now,

b²-4ac=0

{-2(m-1)}²-4*m*(m+2)=0

4(m²+1-2m)-{4m(m+2)}=0

4m²+4-8m-(4m²+8m)=0

4m²+4-8m-4m²-8m=0

4m²-4m²-8m-8m+4=0

0-16m+4=0

-16m=-4

m=-4/-16

m=1/4

Hence,m=1/4 is answer