# Find the value of K for which the roots are real & equal in each of the following equation: (k + 1)x^2 - 2(k?

Find the value of K for which the roots are real & equal in each of the following equation:

(k + 1)x^2 - 2(k - 1)x + 1 = 0

plz help me out show method

### 3 Answers

- 9 years agoBest Answer
[-2(k-1)]^2-4*(k+1)*(1)=0

4k^2+4-8k-4k-1=0

4k^2-12k+3=0

4k^2-6k-6k+3=0

2k(2k-3)-2(2k-3)=0

(2k-2)(2k-3)=0

2k-2=0 i.e. k=2

or

(2k-3)=0 i.e. k+3/2

so...2 values of k.

- warburtonLv 43 years ago
First, improve the unique equation: x^2 + kx - 2x - 2k = 0 (x + ok)(x - 2) = 0 the two (x + ok) = 0 or (x - 2) = 0 x + ok = 0 x = -ok ok = -x x - 2 = 0 x = 2 by way of fact it has actual and equivalent roots: x + ok = x - 2 <-- Subtract x from the two section ok = -2 unsure in case you have been instantaneous by way of your instructor to apply the discriminant... if so: (ok - 2)^2 = -8k (ok - 2)(ok - 2) = -8k ok^2 - 4k + 4 = -8k <-- upload 8k to the two section ok^2 + 4k + 4 = 0 (ok + 2)(ok + 2) = 0 ok + 2 would desire to equivalent 0 to fulfill this equation: ok + 2 = 0 ok = -2 stable success and that i wish I helped!

- 9 years ago
for real and equal roots b^2-4ac=o

From question:

[-2(k-1)]^2-4*(k+1)*(1)=0

4k^2+4-8k-4k-1=0

4k^2-12k+3=0

4k^2-6k-6k+3=0

2k(2k-3)-2(2k-3)=0

(2k-2)(2k-3)=0

2k-2=0 i.e. k=2

or

(2k-3)=0 i.e. k+3/2

so...2 values of k.

Hope that will help...

Source(s): my brain...