Question

Q.2
Find the range of values of a, such that f(x) = ax² + 2(a +1)x+ 9a + 4
is always negative.​

Answer 1

Answer:

Step-by-step explanation:

BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST

Answer 2

Step-by-step explanation:

Considering denominator x

2

−8x+32

D<0 and a>0

So denominator is always positive

For f(x) to be negative, numerator ⇒ax

2

+2(a+1)x+9a+4<0 i.e a<0 and b

2

−4ac<0

⇒ax

2

+2(a+1)x+9a+4<0

⇒a<0 & 4(a+1)

2

−4a(9a+4)<0

⇒4(a

2

+2a+1−9a

2

−4a)<0

⇒4(−8

2

−2a+1)<0

8a

2

+2a−1>0

(4a−1)(2a+1)>0

a>

4

1

or a<−

2

1

but a<0

∴a∈(−∞,−

2

1

)