Question

find the value of k if the roots of the equation are equal x²+2(k+1)x+k²=0​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ x² + 2(k + 1)x + k² = 0.

As we know that,

D = Discriminant  Or  b² - 4ac.

⇒ D = 0  Or  b² - 4ac = 0.

⇒ [2(k + 1)]² - 4(1)(k²) = 0.

⇒ 4(k² + 1 + 2k) - 4k² = 0.

⇒ 4k² + 4 + 8k - 4k² = 0.

⇒ 4 + 8k = 0.

⇒ 8k = -4.

⇒ k = -4/8.

⇒ k = -1/2.

                                                                                                                         

MORE INFORMATION.

Nature of the factors of the quadratic expression.

(1) = Real and different, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal or complex conjugate.

Answer 2

Answer:

k = -1/2

Step-by-step explanation:

Question:

Find the value of k if the roots of the equation are equal x²+2(k+1)x+k²=0​

Solution:

As the given quadratic equation has real and equal roots, then the Discriminant is equal to 0

• D = b^2-4ac = 0

In the given quadratic equation,

• a = 1
• b = 2(k+1) = 2k+2
• c = k^2

Apply the values:

D = (2k+2)^2-4(1)(k^2) = 0

4(k+1)^2 - 4k^2 = 0

4(k^2+2k+1) - 4k^2 = 0

4k^2 + 8k + 4 - 4k^2 = 0

8k+4 = 0

8k = -4

k = -4/8

k = -1/2

Conclusion:

Hence, the value of k is -1/2