Question

If the equation x^2-8x+k=0 has real and distinct roots,then find the value of k.​

Answer 1

kx^2–8x+k=0 be factorised into real linear factor? ... If the difference between the root of the quadratic equation x^2+kx+12=0 is 1, what is

Answer 2

Answer:

Value of k is always less than 16

Step-by-step explanation:

Given equation is

${x}^{2} - 8x + k = 0$

This is a quadretic equation of second degree.

Comparing this equation with

$a {x}^{2} + bx + c = 0$

We get a= 1,b=-8 and c=k

Given condition is roots are real and distinct

So,

${b}^{2} - 4ac > 0$

Putting value of a,b,c and get,

$( { - 8})^{2} - (4 \times 1 \times k) > 0$

$64 - 4k > 0$

$64 > 4k$

$k < 16$

Here value of k is always less than 16.