For what values of k does the quadratic equation (k+3)x^2 + 4x + k = 0 have no real roots.

Question

Please state and show how to answer the question and explain how to do it, thank you! :)

Answer 1

The discriminant of Ax²+Bx+C is B²-4AC

 

If the discriminant is negative, then there are no real roots.

 

So, for (k+3)x² + 4x + k = 0, we want 4² - 4(k+3)(k) < 0

 

                                                         16 - 4(k²+3k) < 0

 

                                                         -4k² - 12k + 16 < 0

 

                                                            k² + 3k - 4 > 0

 

                                                           (k+4)(k-1) > 0

 

(k+4)(k-1) = 0 when k = -4 or 1

 

If k < -4, then (k+4)(k-1) > 0

 

If -4 < k < 1, then (k+4)(k-1) < 0

 

If k > 1, then (k+4)(k-1) > 0

 

So, there are no real roots when k < -4 or k > 1.