Question

Find the values of ‘k’ such that quadratic equation (k* + 9)x2 + (k +1) x +1 = 0 has no real roots?

Answer 1

Answer:

k∈(-5,7).

Step-by-step explanation:

If a quadratic equation does not have any real roots then its discriminant must be less than zero.

i.eΔ<0

GIven quadratic equation is

(k+9)x² + (k+1)x +1 = 0

Its discriminant is,

(k+1)² - 4(k+9)(1)  (∵Δ=b²-4ac)

It must be less than zero,

i.e k² + 1 + 2k -4k -36 <0

⇒k²-2k-35<0

⇒k²-7k+5k-35<0

⇒k(k-7) + 5(k-7)<0

⇒(k+5)(k-7)<0

If you observe if k<-5 then the term above is positive (∵ - x - =+).

It is also positive if k>7.

Hence k must lie in between -5 and 7, i.e k∈(-5,7).

Answer 2

Answer:

ATQ,

D=B²-4AC

K²+2K+1-4K-36<0

K²-2K-35<0

[K+5][K-7]<0

K=-5,7.

Hence k must lie between -5 ,7.