Question

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

Answer 1

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.

Answer 2

Answer:

Given ,

Quadratic equations

${x}^{2} \: + \: 2x \: \: + k \: = 0$

We know that;

️‍️

D=Discriminat

or,

${b}^{2} \: - 4ac = 0$

Roots are real and equal

therefore, D = 0

Now,

=>

${2}^{2} \: - \: 4(1)(k) = 0$

$=> \: 4 - 4k \: = 0$

Step-by-step explanation:

$=> 4 = 4k$

$=> k = 1$

️‍️

therefore, K= 1 is the answer