Question

Find the value of k for which the quadratic equation is 3x²-5x+2k has real and equal roots

Answer 1

if a equation has real and equal root (b^2-4ac=0)

(-5)^2-4(3)(2k) =0
25-24k=0
k=25/24 is answer

Answer 2

Question:

Find the value of k for which the quadratic equation 3x² - 5x + 2k = 0 has equal roots.

Answer:

k = 25/24

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 .

• 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 ;

3x² - 5x + 2k = 0

Clearly , we have ;

a = 3

b = -5

c = 2k

We know that ,

The quadratic equation will have equal roots if its discriminant is equal to zero .

=> D = 0

=> (-5)² - 4•3•2k = 0

=> 25 - 24k = 0

=> 24k = 25

=> k = 25/24

Hence,

The required values of k is 25/24 .