# What is a if the equation has real roots? ax^2+(3a-1)x+a+3=0

neela | Student

What is a if the equation has real roots?

ax^2+(3a-1)x+a+3=0.

The discriminant of Ax^2+Bx+C = B^2-4AC.

If the discriminant is positive then Ax^2+Bx+c = 0 has real roots.

So the discriminant of the quadratic D =  (3a-1)^2 - 4a(a+3)

D = (3a-1)^2 - 4a(a+3)

D = 9a^2-6a+1 - 4a^2-12a

D = 5a^2 -18a +1.

D = 5(a^2 -18/5)+1.

D = 5(a-9/5)^2 - 81/5+1

D=  5(a-9/5)^2 -76/5 > 0 if 5(a-9/5)^2 > 76/5.

(a-9/5)^2 > 76/25

a -9/5 > sqrt(76/5), Or a> 9/5+sqrt(76/)/5.

a -9/5 < -sqrt(76)/5.

a < 5/9 - sqrt(76)/5.

Therefore the equation ax^2+(3a-1)x+a+3=0 has real roots if a> 9/5+sqrt(76/)/5 or a < 5/9 - sqrt(76)/5.

giorgiana1976 | Student

For the given equation to have real roots, the discriminant delta has to be positive or zero.

delta = (3a-1)^2 - 4a(a+3)

delta >= 0

We'll expand the square and we'll remove the brackets in the expression of delta:

9a^2 - 6a + 1 - 4a^2 - 12a >= 0

We'll combine like terms:

5a^2 - 18a + 1 >= 0

We'll have to determine first the roots of the expression 5a^2 - 18a + 1:

5a^2 - 18a + 1 = 0

a1 = [18+sqrt(324 - 20)]/10

a1 = (18+4sqrt19)/10

a1 = (9+2sqrt19)/5

a2 = (9-2sqrt19)/5

The expression 5a^2 - 18a + 1 >= 0 for the values of a from the intervals:

(-infinite ; (9-2sqrt19)/5] U [(9+2sqrt19)/5 ; +infinite)