1.Find values of k which the equation (2-k)x^2-4(k-2)x-5=0, has two real roots.Thankyou for answwering in advance. Please show steps carefully. ALSO PLEASE EXPLAIN TO ME WHAT DERIVATIVES ARE.
A quadratic function, `y=ax^2+bx+c`, has two real roots if the discriminant,`b^2-4ac >0` .(If the discriminant is less than zero, there are no real roots, but there are two imaginary roots.If the discriminant is zero, there is one real root.)
(1) Given `y=(2-k)x^2-4(k-2)x-5` we have `a=2-k,b=-4(k-2),c=-5`
(2) The discriminant is `[-4(k-2)]^2-4(2-k)(-5)=16k^2-84k+104`
(3) The function has two real roots if the discriminant is greater than zero:
Then the discriminant is zero at `k=2,k=13/4`
`16k^2-84k+104` represents the graph of a parabola that opens up. The function is greater than zero on `(-oo,2)uu(13/4,oo)` .
Thus if `k<2` or `k>13/4` the function `y=(2-k)x^2-4(k-2)x-5` will have two real zeros.
Example graphs with k=1(red),k=3(blue),and k=5(green)
As far as a derivative goes, to explain even partially requires some knowledge of limits. It typically takes 1/4 of a college semester to explain the basics of the derivative. In short, it is the instantaneous rate of change of a function at a point.