Find the values of k for which the quadratic equation
(3k + 1)x² + 2(k + 1)x + 1 = 0 has equal roots. Also find the roots.
Answers
Given:
The quadratic equation (3k + 1)x² + 2(K + 1)x + 1 = 0 has equal roots.
Find:
The value of k.
Solution:
The given quadratic equation (3k + 1)x² + 2(K + 1)x + 1 = 0 and root are real and equal.
Here,
a = 3k + 1
b = 2(k + 1)
c = 1
We know that:
• D => b² - 4ac
Putting the value of a = 3k + 1, b = 2(k + 1) and c = 1
• D => b² - 4ac
=> [2(k + 1)]² - 4(3k + 1) (1)
=> 4(k² + 2k + 1) - 12k - 4
=> 4k² + 8k + 4 - 12k - 4
=> 4k² - 4k
The given equation will have real and equal roots,
• D = 0
Thus,
=> 4k² - 4k = 0
=> 4k(k - 1) = 0
=> k = 0 Or k - 1 = 0
=> k = 0 Or k = 1
Therefore,
The value of k is o or 1.
Now,
for k = 0, the equation.
=> x² + 2x + 1 = 0
=> x² + x + x + 1 = 0
=> x(x + 1) + 1(x + 1) = 0
=> x = -1 , -1
for k = 1, the equation.
=> 4x² + 4x + 1 = 0
=> 4x² + 2x + 2x + 1 = 0
=> 2x(2x + 1) + 1(2x + 1) = 0
=> (2x + 1)² = 0
=> x = -1/2 , -1/2
Hence, the roots of the equation are -1 and -1/2
I hope it will help you.
Regards.
Answer:
• D = 0
Thus,
=> 4k² - 4k = 0
=> 4k(k - 1) = 0
=> k = 0 Or k - 1 = 0
=> k = 0 Or k = 1
Therefore,
The value of k is o or 1.
Now,
for k = 0, the equation.
=> x² + 2x + 1 = 0
=> x² + x + x + 1 = 0
=> x(x + 1) + 1(x + 1) =
Step-by-step explanation: