Question

15. If the roots of then equation x + 2(a +1)x+ 9a - 5 = 0 are real and unequal then the
Value of a is
a) 1<a<6 b) 1<_ a <_ 6
c) a< 1 , a > 6
d) a<_ 1 , a >_ 6​

Answer 1

Step-by-step explanation:

If a=0, it becomes linear equation.

If b

2

−4ac=0, then there will be real and equal roots.

If b

2

−4ac<0, then the roots will be unreal.

Only if b

2

−4ac>0, we will get two real distinct roots.

Option D is correct!

Answer 2

Answer:

Given that roots are real and unequal

So, the Discriminant (∆) > 0

Given that quadratic equation :

→ x² + 2(a+1) x + 9a -5 = 0

Discriminant (∆) = b² - 4ac

4(a+1)² - 4(9a-5) > 0

4a² + 4 + 8a -36a + 20 > 0

4a² - 28a + 24 > 0

4a² - 24a - 4a + 24 >0

4a(a-6) - 4(a-6) > 0

(a-6) (4a-4) > 0

For the value must be greater than 0

a - 6 > 0 4a - 4 > 0

a > 6 a > 1

Hope it's helpful to all