Question

The value of k for equation 9xsquare- 8xk+8=0 has equal roots

Answer 1

Question:

Find the valu of "k" for equation

9x^2 - 8kx + 8 = 0, has equal roots .

Answer:

k = ±3/√2 or ±3√2/2

Notes:

• The degree of equations decides the number of its roots.

• The maximum number of roots of an equation is equal to its degree.

• The degree of a quadratic equation is two, thus it will have maximum two roots.

• If we have a quadratic equation

ax^2 + bx + c = 0 , then its determinant is given by , D = b^2 - 4•a•c

• If D = 0 , then the quadratic equation will have real and equal roots.

• If D > 0 , then the quadratic equation will have real and distinct roots.

• If D < 0 , then the quadratic equation will have non-real(imaginary) roots.

Solution:

The given quadratic equation is ;

9x^2 - 8kx + 8 = 0.

Also,

The determinant for the given equation will be given as ;

=> D = (-8k)^2 - 4•9•8

=> D = 32(2k^2 - 9)

We know that;

For equal roots, the determinant of the given quadratic equation must be zero.

Thus,

=> D = 0

=> 32(2k^2 - 9) = 0

=> 2k^2 - 9 = 0

=> 2k^2 = 9

=> k^2 = 9/2

=> k = ±√(9/2)

=> k = ± 3/√2

=> k = ±3√2/2

Hence,

The required value of "k" are ±3/√2 or

± 3√2/2 .