Multiple choice

Let $a, b$ and $c$ be three real numbers, such that $a+2b+4c=0$. Then the equation $a{ x }^{ 2 }+bx+c=0$

  1. has both the roots complex
  2. has its roots lying within $-1< x< 0$
  3. has one of the roots equal to $\cfrac{1}{2}$
  4. has its roots lying within $2< x< 6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given a + 2b + 4c = 0, we can write c = -a/4 - b/2. Substituting this into the quadratic equation ax^2 + bx + c = 0 gives ax^2 + bx - (a/4 + b/2) = 0. Testing x = 1/2: a(1/4) + b(1/2) - a/4 - b/2 = 0, which confirms x = 1/2 is a root.

AI explanation

Since a + 2b + 4c = 0, substituting x = 1/2 into the given equation yields a(1/2)^2 + b(1/2) + c = 0, which simplifies to (a + 2b + 4c)/4 = 0. Because the numerator perfectly matches the given condition, the expression evaluates to zero. Therefore, x = 1/2 is a root of the equation ax^2 + bx + c = 0.

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