Question

hey guys plzzzzzz ans this one briefly with process plzzz​

Answer 1

Answer:

Step-by-step explanation:

For an equation ax²+bx+c = 0, the roots are real and equal when the discriminant i.e. b²- 4ac is equal to 0.

So in the given equation, (2k+1)x² + 2(k+3)x + (k+5) = 0

a = 2k+ 1 , b = 2(k+3), c = k+5

For roots to be real and equal, b² - 4ac = 0

=> [2(k+3)]² - 4(2k+ 1)(k+5) = 0

=> 4(k+3)² - 4(2k² + 10k + k + 5) = 0

=> 4 [ k² + 6k + 9 - 2k² - 11k -5] = 0

=> 4[ -k² - 5k + 4] = 0

=> k² + 5k - 4 = 0

We know that for quadratic equation of form ax² + bx + c = 0,

x = [-b ±√(b² - 4ac)] / 2a

Thus value of k = {-5 ±√[(5²) - 4(1)(-4)] } / 2(1)

=> k = [-5 ± √41] / 2

Thus the equation has real and equal roots when

k = [-5+√41] / 2  or  k = [-5 -√41] / 2