Multiple choice

Let $f\left( x \right) =3a{ x }^{ 2 }-4bx+c\left( a,b,c\in R,a\neq 0 \right) $ where $a,\ b,\ c$ are in A.P. Then the equation $f\left( x \right) =0$ has

  1. no real solution

  2. two unequal real roots

  3. sum of roots always negative.

  4. product of roots always positive.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

a, b, c in A.P. means 2b = a + c. f(x) = 3ax^2 - 4bx + c = 3ax^2 - 2(a+c)x + c = 0. Discriminant D = 4(a+c)^2 - 4(3a)(c) = 4(a^2 + 2ac + c^2 - 3ac) = 4(a^2 - ac + c^2). Since a^2 - ac + c^2 is always positive for a, c != 0, the roots are real and unequal.

AI explanation

Using the discriminant of 3a x^2 - 4bx + c = 0, we calculate (-4b)^2 - 4(3a)(c). This simplifies to 16b^2 - 12ac. Since a, b, c are in an arithmetic progression, 2b = a + c, so b = (a+c)/2. Substituting this into the discriminant gives 16((a+c)/2)^2 - 12ac = 4a^2 - 4ac + 4c^2 = 4(a^2 - ac + c^2). Since a^2 - ac + c^2 is always positive for real a and c, the discriminant is always positive. Thus, the equation has two unequal real roots.

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