Question

find tge value of k for which the quadratic equation kx square-5x+k have real roots​

Answer 1

Step-by-step explanation:

• b square- 4ac = 0
• a = k. b = -5. c = k
• (-5)square - 4 * k*k = 0
• 25 - 4k square = 0
• - 4k square =- 25(minus minus cancelled)
• 4k = under root 25
• 4k = 5
• k = 5/4

Answer 2

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] .