Question

Find the value(s) of k for which the equation x squre+5kx+16=0 has real and equal roots. .

Answer 1

Discriminant = b^2 - 4ac

(5k)^2 - 4×1 × 16 =0 ( D= 0)

25k^2 - 64 = 0

K^2 = 64 ÷ 25

K = root 64/ 25

K = 8 / 5 = 1.6

Answer 2

Answer:

The required value of k is 8 / 5.

Step-by-step explanation:

Given equation is x^2 + 5kx + 16 = 0, also , given that this equation has real and equal roots.

On comparing this equation with ax^2 + bx + c = 0, we can say that the value of a is 1 , b is 5k and c is 16 .

From the properties of quadratic equations, we know that if an equation has equal and real roots, value of its discriminant is 0.

Therefore,

= > Discriminant = 0

= > b^2 - 4ac = 0

= > ( 5k )^2 - 4( 1 )( 16 ) = 0

= > ( 5k )^2 - 64 = 0

= > ( 5k )^2 - ( 8 )^2 = 0

= > ( 5k )^2 = ( 8 )^2

= > 5k = 8

= > k = 8 / 5

Hence the required value of k is 8 / 5.