Determine the value of k for which the function f(x)= 4x^2- 3x+2kx+1 has two zeros. check these values in the original equation?

Question

i cant figure out the answer to it. It says in the book that it is k<-0.5 or k> 3.5,but i don't know how they got it.

Answer 1

The zeros of a function are determined by the roots of the corresponding equation.
So let's look at
4x^2 - 3x + 2kx + 1 = 0
4x^2 + x(2k-3) + 1 = 0 , where a=4, b=2k-3 and c=1

I will assume you know the quadratic formula.
The nature of the roots depends on the value of the
b^2 - 4ac part of the formula, the part inside the √ sign.

if b^2 - 4ac = 0 there will be 1 real root
if b^2 - 4ac < 0 there will be no real roots,
if b^2 - 4ac > 0 there will be 2 different real roots

so we want the last part

b^2 - 4ac > 0
(2k-3)^2 - 4(4)(1) > 0
4k^2 - 12k + 9 - 16 > 0
4k^2 - 12l - 7 > 0
(2k + 1)(2k - 7) > 0

I this had been an equation then its roots would have been
k = -1/2 = -.5 or k = 7/2 = 3.5

so k < -.5 or k > 3.5

Answer 2

Heyy just wanted to point out where it says "12l" it's suppose to be 12k , the calculation is fine, just a simple typo. :)