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