Question

find m, if quadratic equation (m-1)x²-2(m-1)x+1=0 has real and equal values​

Answer 1

Step-by-step explanation:

(m-1)x²-2(m-1)x+1=0

roots are real and equal if D=0

=> b^2-4ac = 0

=> {-2(m-1)}^2 - 4(m-1)1 = 0

=> 4(m-1)^2-4(m-1) = 0

=> 4(m-1)(m-1-1) = 0

=> (m-1)(m-2) = 0

=> m-1 = 0 or m-2 = 0

=> m = 1 or 2

Answer 2

Answer:

The value of m is 1 or 2

Step-by-step explanation:

The general form of the quadratic equation is $ax^{2} + bx + c$

To find the roots of quadratic equation by Shridhar's method is given by

x = (-b ±$\sqrt{b^{2}-4ac }$ )/ 2a

For the roots to be equal and real ${b^{2}-4ac }= 0$  

Given quadratic equation $(m-1)x^{2} -2(m-1)x+1$

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

Given that roots are real and equal therefore

${b^{2}-4ac }= 0$

$(-2(m-1))^{2}-4(m-1)(1) = 0$

$4(m-1)^{2} - 4 (m-1)= 0$

$4(m-1)(m-2) = 0$

Therefore m = 1 or m = 2.