Question

for all x^2 , x^2 + 2ax +(10-3a) >0 , then the interval in which 'a' lies is ?fasttt​

Answer 1

D<0⇒4

a 2

−4(10−3a)<0⇒4

a 2

+12a−40<0⇒−5<a<2.

D<0⇒4a2−4(10−3a)<0⇒4a2+12a−40<0⇒−5<a<2.

Answer 2

Step-by-step explanation:

b^2-4ac<0

4a^2-4(1)(10-3a)<0

4a^2-40+12a<0

a^2+3a-10<0

a^2+5a-2a-10<0

a(a+5)-2(a+5)<0

(a-2)(a+5)<0

a belongs to (-5,2)