3. x2 + 2x + k = 0 If the roots of the given equation are real
and equal then find k.​

2 thoughts on “3. x2 + 2x + k = 0 If the roots of the given equation are real<br />and equal then find k.​”

1. Answer:

   Given ,

   Quadratic equations

   [tex] {x}^{2} \: + \: 2x \: \: + k \: = 0 \\ [/tex]

   We know that;

   ️‍️

   D=Discriminat

   or,

   [tex] {b}^{2} \: – 4ac = 0[/tex]

   Roots are real and equal

   therefore, D = 0

   Now,

   =>

   [tex] {2}^{2} \: – \: 4(1)(k) = 0[/tex]

   [tex]=> \: 4 – 4k \: = 0[/tex]

   Step-by-step explanation:

   [tex]=> 4 = 4k[/tex]

   [tex]=> k = 1[/tex]

   ️‍️

   therefore, K= 1 is the answer

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2. EXPLANATION.

   Quadratic equation.

   ⇒ x² + 2x + k = 0.

   As we know that,

   D = Discriminant or b² – 4ac = 0.

   Roots are real and equal, D = 0.

   ⇒ (2)² – 4(1)(k) = 0.

   ⇒ 4 – 4k = 0.

   ⇒ 4 = 4k.

   ⇒ k = 1.

   MORE INFORMATION.

   Conjugate roots.

   (1) = D < 0.

   One roots = α + iβ.

   Other roots = α – iβ.

   (2) = D > 0.

   One roots = α + √β.

   Other roots = α – √β.

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