The given equation 4x2– 2(k +1)x + (k + 4)= 0 is in the form of ax2+ bx + c = 0
Where a = 4, b = -2(k + 1), c = (k + 4)
For the equation to have real and equal roots, the condition is
D = b2– 4ac = 0
⇒ (-2(k + 1))2– 4(4)(k + 4) = 0
⇒ 4(k +1)2– 16(k + 4) = 0
⇒ (k + 1)2– 4(k + 4) = 0
⇒ k2– 2k – 15 = 0
Now, solving for k by factorization we have
⇒ k2– 5k + 3k – 15 = 0
⇒ k(k – 5) + 3(k – 5) = 0
⇒ (k – 5)(k + 3) = 0,
k = 5 and k = -3,
So, the value of k can either be 5 or -3.
