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

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### 2 Answers

A quadratic equation ax^2 + bx + c = 0 has real roots if b^2 >= 4ac

Here the equation is ax^2+(3a-1)x+a+3=0

(3a - 1)^2 > = 4*(a + 3)*a

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

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

For this to happen, if we can represent the expression as (x - x1)(x - x2) >= 0, either both should be positive or both negative. This is true for values which do not lie between x1 and x2

x1 = 18/10 + sqrt( 324 - 20)/10

=> x1= 1.8 + sqrt (304) /10

x2 = 1.8 - sqrt (304) / 10

**This gives the interval that a lies in as (-inf. , 1.8 - (sqrt 304)/10) U (1.8 + (sqrt 304)/10) , +inf.)**

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

We'll apply the quadratic formula:

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)**