QUADRATIC EQUATION HELP

Question

Find the range of values of k for which the equation 2x²+x+5=k(x+1)² has no real roots.

Answer 1

2x² + x + 5 = k(x² + 2x + 1)

 

(2-k)x² + (1-2k)x + (5-k) = 0

 

Ax² + Bx + C = 0 has no real roots when the discriminant, B² - 4AC, is negative.  

 

So, we need to find k so that (1-2k)²-4(2-k)(5-k) < 0

 

1 - 4k + 4k² - 4(10 - 7k + k²) < 0

 

1 - 4k + 4k² - 40 + 28k - 4k² < 0

 

24k - 39 < 0

 

k < 39/24

 

So, k < 13/8

 

Or, equivalently, k < 1.625