Question

For what values of 'a' quadratic equation 3 ax^2- 6x + 1 = has no real roots​

Answer 1

For no real roots

b^2 - 4ac < 0

36 - 12a < 0

12(3-a) < 0

3-a < 0

3 < a

a > 3

It mean the value of a is greater than 3.

Answer 2

Answer:

$\huge\boxed{\text{for}\ a>3}$

Step-by-step explanation:

For $ax^2+bx+c=0$

if

$b^2-4ac<0$ then te eqution has no real solution

We have:

$3ax^2-6a+1=0\\\\b^2-4ac=(-6)^2-4(3a)(1)=36-12a\\\\36-12a<0\\-36\qquad-36\\\\-12a<-36\qquad|\text{change the signs}\\\\12a>36\\\\\dfrac{12a}{12}>\dfrac{36}{12}\\\\a>3$