find k. x²-2[2k-1]x+k²=0 ha real and equal roots
Answers
Answer:
k= ?
Step-by-step explanation:
Given x
2
−2(k+1)x+k
2
=0 has real and equal roots
As we know that
for a quadratic equation to have real and equal roots ,discriminant should be zero
⟹(−2(k+1)) 2
−4(1)(k 2 )=0
⟹(k+1)
2 −k 2 =0
⟹2k+1=0
⟹k=−1/2
hope it helps you and other students
Answer:
The values of k are 1 and 1 / 3.
Step-by-step-explanation:
The given quadratic equation is x² - 2 ( 2k - 1 ) x + k² = 0.
We have given that,
The equation has real and equal roots.
Comparing given equation with ax² + bx + c = 0, we get,
- a = 1
- b = - 2 ( 2k - 1 ) = - 4k + 2
- c = k²
For real and equal roots,
b² - 4ac = 0
⇒ ( - 4k + 2 )² - 4 * 1 * k² = 0
⇒ ( - 4k )² - 2 * 4k * 2 + 2² - 4k² = 0
⇒ 16k² - 16k + 4 - 4k² = 0
⇒ 16k² - 4k² - 16k + 4 = 0
⇒ 12k² - 16k + 4 = 0
⇒ 4 ( 3k² - 4k + 1 ) = 0
⇒ 3k² - 4k + 1 = 0
⇒ 3k² - 3k - k + 1 = 0
⇒ 3k ( k - 1 ) - 1 ( k - 1 ) = 0
⇒ ( k - 1 ) ( 3k - 1 ) = 0
⇒ ( k - 1 ) = 0 OR ( 3k - 1 ) = 0
⇒ k - 1 = 0 OR 3k - 1 = 0
⇒ k = 1 OR 3k = 1
⇒ k = 1 OR k = 1 / 3
∴ The values of k are 1 and 1 / 3.