Question

Marks: 3
Without solving the following quadratic equation find the value of m for which the
given equation has real and equal roots .
X2 +2(m-1) x +(m+5) =0​

Answer 1

Step-by-step explanation:

For real and equal roots discriminant (d) should be 0

therefore, b^2-4ac=0

[2 (m-1)]^2 -4 (m+5)×1 =0

4 (m-1)^2 =4 (m+5 )

(m -1 )^2 = m+5

m^2-2m+1 =m+5

m^2-3m-4 =0

m^2 -4m+m -4 = 0

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

m= 4 or -1

Hope it helps ..please mark it the brainliest

Answer 2

Answer:

(- 1) or 4

Step-by-step explanation:

Equation (ax ± b)² = 0 has real and equal roots.

(a ± b)² = a² ± 2ab + b²

~~~~~~~~~~

x² + 2(m-1)x + (m+5) = 0

(m - 1)² = m + 5

m² - 2m + 1 - m - 5 = 0

m² - 3m - 4 = 0

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

$m_{1}$ = - 1 and $m_{2}$ = 4

1). If m = - 1, then

x² + 2(-1-1)x + (-1+5) = 0 ===> x² - 4x + 4 = (x - 2)² = 0

(x - 2)² = 0 ===> $x_{12}$ = 2

2). If m = 4, then

x² + 2(4-1)x + (4+5) = 0 ===> x² + 6x + 9 = (x + 3)² = 0

(x + 3)² = 0 ===>  $x_{12}$ = - 3