Question

(m-1)x²-2(m - l) x + 1 = 0
--
find the
value of m, if the roots of quadratic equation are real and equal.
-​

Answer 1

Answer :

m = 2

Note:

★ The possible values of the variable which satisfy the equation are called its roots or solutions .

★ A quadratic equation can have atmost two roots .

★ The general form of a quadratic equation is given as ; ax² + bx + c = 0

★ If α and ß are the roots of the quadratic equation ax² + bx + c = 0 , then ;

• Sum of roots , (α + ß) = -b/a

• Product of roots , (αß) = c/a

Solution :

Here ,

The given quadratic equation is ;

(m - 1)x² - 2(m - 1)x + 1 = 0

Comparing the given quadratic equation with the general quadratic equation ax² + bx + c = 0 , we have ;

a = m - 1

b = -2(m - 1)

c = 1

For real and equal roots , the discriminant of the equation must be zero .

Thus ,

=> D = 0

=> b² - 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 , 2

=> m = 2 (Appropriate value)

Note : m = 1 is rejected value . Because if m = 1 , then the equation will not be quadratic .

Hence , m = 2 .

Answer 2

Step-by-step explanation:

Given quadratic equation is (m - 1)x² - 2(m - 1)x + 1 = 0

Also said that the roots of the quadratic equation are real and equal.

We have to find the value of m.

Quadratic formula:

x = -b ± √(b² - 4ac)/2a

OR

x = - b ± √D/2a

D = b² - 4ac

As the roots of the quadratic equation are real and equal. So, "D" must be equal to zero.

→ D = b² - 4ac = 0

→ [-2(m - 1)]² - 4(m - 1)(1) = 0

→ 4(m - 1)² - 4(m - 1) = 0

→ 4(m² + 1 - 2m) - 4(m - 1) = 0

→ 4m² + 4 - 8m - 4m + 4 = 0

→ 4m² - 12m + 8 = 0

→ 4(m² - 3m + 2) = 0

→ m² - 3m + 2 = 0

→ m² - 2m - m + 2 = 0

→ m(m - 2) -1(m - 2) = 0

→ (m - 1)(m - 2) = 0

→ m = 1, 2

(One is neglected. Because if we put the value of m as 1 then equation will not be quadratic equation)

Hence, the value of m is 2.