Please help me with this quastion

Question

Given the quadratic equation (2k+1)x²  - 3kx+2k-4=0 has no roots.Find the range of the values of k

Answer 1

We can use the fact that if a quadratic equation (ax²+bx+c = 0) has no (real) roots, then the discriminant (b² - 4ac) should be negative.

 

In this problem:

a = 2k+1

b = -3k

c = 2k-4

 

discriminant = (-3k)² - 4(2k+1)(2k-4)

 

Simplify.

 

discriminant = -7k² + 24k + 16

 

To determine the intervals, we need to figure out the roots of THIS new quadratic.

 

k₁ = [-24 + sqrt(24² - 4(-7)(16))] / [-14] =  -4/7

k₂ = [-24 - sqrt(242 - 4(-7)(16))] / [-14] = 4

 

So, we've determined the edges of the intervals... And we know if we graphed y = -7x² + 24x + 16, we would have a downwards facing parabola, because the leading coefficient is negative. So,

 

-7k² + 24k + 16 < 0

when

 

-infinity < k < -4/7  or  4 < k < infinity

 

Which, in interval notation gives:

(-inf , -4/7) U (4 , inf) 

Answer 2

NOTE: Every quadratic equation has root, whether real, imaginary, or combination of both.

 

I believe the question wants the range which it does not have any real roots. Also we know if the number under radical (square root) is negative, then it does not have REAL answer. Set the following equation (for more details refer to your book)

 

(-3k)²-4*(2k+1)(2k-4)=9k²-4(4k²-6k-4)=9k²-16k²+24k+16=-7k²+24k+16 if this is less than zero, then your original equation does not have

-7k²+24k+16<0 which is another quadratic equation. Solving this equation for k gives the range you need!