(1) If the roots of x2 + kx + k = 0 are real and equal, what is the value of k?
(A) 0
(B) 4 (C) 0 or 4
(D) 2​

2 thoughts on “(1) If the roots of x2 + kx + k = 0 are real and equal, what is the value of k?<br />(A) 0<br />(B) 4 (C) 0 or 4<br />(D) 2​”

1. Answer:

   Given :–

   [tex] \bf \: {x}^{2} + kx + k[/tex]

   To Find :–

   Value of k

   Solution :–

   Since there root are real and equal

   D = b² – 4ac

   [Discriminant Rule]

   D = 0

   b = k

   a = 1

   c = k

   0 = k² – 4(1)(k)

   0 = k² – 4k

   0 = k(k – 4)

   Either

   [tex] \sf \: k = 0 + 4[/tex]

   [tex] \sf \: k = 4[/tex]

   Or

   [tex] \sf \: k = 0[/tex]

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2. Answer:

   0 or 4

   Step-by-step explanation:

   To roots to be real and equal, discriminant of the equation must be 0.

   Discriminant of ax² + bx + c = 0 is given by b² – 4ac. On comparing,

   a = 1, b = k, c = k

   ⇒ discriminant = 0

   ⇒ k² – 4(1)(k) = 0

   ⇒ k² – 4k = 0

   ⇒ k(k – 4) = 0

   ⇒ k = 0 or k – 4 = 0

   ⇒ k = 0 or k = 4

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