9.
For what value(s) of k, the quadratic equation x²-2(k+1)x+k²=0 has equal roots
Answers
EXPLANATION.
Quadratic equation.
⇒ x² - 2(k + 1)x + k² = 0.
As we know that,
D = Discriminant Or b² - 4ac.
⇒ [-2(k + 1)²] - 4(1)(k²) = 0.
⇒ 4(k + 1)² - 4k² = 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:
Given
- x² - 2(k + 1)x + k² = 0
To find
- The value of k for which the quadratic equation has equal roots.
Solution
The discriminant of a quadratic equation tells us whether there are two solutions, one solution, or no solutions.
Discriminant = b² - 4ac = 0
(Since they are equal roots, it is equal to 0)
Quadratic formula = ax² + bx + c = 0
Where: a = 1 , b = 2 (k + 1) , c = k²
- (-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
Hence, the value of k is -1/2