find the value of k for which the quadratic equation k square-5x+k=o have real roots
Answers
Question:
Find the value of k for which the quadratic equation kx² - 5x + k = 0 has real roots.
Answer:
k € [-5/2 , 5/2]
Note:
• An equation of degree 2 is know as quadratic equation .
• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.
• The maximum number of roots of an equation will be equal to its degree.
• A quadratic equation has atmost two roots.
• The general form of a quadratic equation is given as , ax² + bx + c = 0 .
• A quadratic equation has atmost two roots .
• The discriminant of the quadratic equation is given as , D = b² - 4ac .
• If D = 0 , then the quadratic equation would have real and equal roots .
• If D > 0 , then the quadratic equation would have real and distinct roots .
• If D < 0 , then the quadratic equation would have imaginary roots .
Solution:
The given quadratic equation is ;
kx² - 5x + k = 0
Clearly , we have ;
a = k
b = -5
c = k
We know that ,
The quadratic equation will have real roots if its discriminant is greater than or equal to zero .
=> D ≥ 0
=> (-5)² - 4•k•k ≥ 0
=> 25 - 4k² ≥ 0
=> 25/4 - k² ≥ 0
=> k² - 25/4 ≤ 0
=> (k-5/2)(k+5/2) ≤ 0
=> k € [-5/2 , 5/2]
Hence,
The required values of k are [-5/2 , 5/2] .
Answer:
The solution set of values of k is
k ∈ [ - 5 / 2, 5 / 2 ].
Step-by-step-explanation:
The given quadratic equation is
kx² - 5x + k = 0.
Comparing with ax² + bx + c = 0, we get,
- a = k
- b = - 5
- c = k
For real roots,
b² - 4ac ≥ 0
⇒ ( - 5 )² - 4 * k * k ≥ 0
⇒ 25 - 4k² ≥ 0
⇒ 25 ≥ 4k²
⇒ 25 / 4 ≥ k²
⇒ k² ≤ 25 / 4
⇒ k ≤ ± 5 / 2
⇒ k ≤ 5 / 2 OR k ≥ - 5 / 2
∴ - 5 / 2 ≤ k ≤ 5 / 2
∴ The solution set of values of k is
k ∈ [ - 5 / 2, 5 / 2 ].