Question

x^+k[4x+k-1]+2=0 has real n equal roots so find the value of k

Answer 1

Hi Mate!!!

For real and equal roots Descrimnant ( D ) must be =0

D = √ { k² - 8 }

D = 0

√ ( k² - 8 ) = 0

squaring both sides.

k² = 8

k = 2√2 or k = -2√2

Answer 2

SOLUTION :  

Option (b) is correct :  ⅔ , - 1

Given : x² + k(4x + k - 1)  + 2 = 0

x² + 4kx + k² - k + 2 = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = 1 , b =  4k  , c = k² - k + 2

D(discriminant) = b² – 4ac

D = (4k)² - 4 × 1 × (k² - k + 2)

D = 16k² - 4k² + 4k - 8

D = 12k² + 4k - 8

D = 4(3k² + k - 2)

D = 0 (equal roots given)

4(3k² + k - 2) = 0

(3k² + k - 2) = 0

3k² + 3k - 2k - 2 = 0

[By middle term splitting]

3k (k + 1) - 2(k + 1) = 0

(3k - 2)(k + 1) = 0

(3k - 2) = 0 (k + 1) = 0

3k = 2 or k = - 1

k = ⅔ or k = - 1

Hence, the value of k is ⅔ & - 1.

HOPE THIS ANSWER WILL HELP YOU...