# Find the value of k = ( x^2 - 4 )/ ( 2x -5 ) if the roots of the equation are equal .

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have k = (x^2 - 4)/( 2x - 5)

k = (x^2 - 4)/( 2x - 5)

=> x^2 - 4 = 2kx - 5k

=> x^2 - 2kx + 5k - 4 = 0

As the roots of the quadratic equation are equal, b^2 - 4ac = 0

=> (-2k)^2 - 4*( 5k - 4) = 0

=> 4k^2 - 20k + 16 = 0

=> k^2 - 5k + 4 = 0

=> k^2 - 4k - k + 4 = 0

=> k(k - 4) - 1(k - 4) = 0

=> (k-1)(k-4) = 0

=> k = 1 and k = 4

Therefore the values of k are 1 and 4.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll multiply both sides by 2x - 5.

k(2x - 5) = (x^2 - 4)(2x - 5)/ (2x - 5)

We'll simplify and we'll get:

k(2x - 5) = (x^2 - 4)

We'll remove the brackets:

2kx - 5k = x^2 - 4

We'll move all terms to one side:

x^2 - 4 - 2kx + 5k = 0

We'll combine like terms:

x^2 - 2kx + 5k - 4 = 0

For the roots of the quadratic to be equal, the discriminant delta has to be zero.

delta = b^2 - 4ac

a,b,c are the coefficients of the quadratic.

delta = (-2k)^2 - 4(5k - 4)

delta = 4k^2 - 20k + 16

4k^2 - 20k + 16 = 0

We'll divide by 4:

k^2 - 5k + 4 = 0

k1 = [5 + sqrt(25 - 16)]/2

k1 = (5+3)/2

k1 = 4

k2 = (5-3)/2

k2 = 1

The values of k, for the equation to have equal roots, are: {1 ; 4}.