Multiple choice

Let $a,b$ and $c$ be real numbers such that $4a+2b+c=0$ and $ab>0$.then the equation ${ ax }^{ 2 }+bx+c=0$ has

  1. Real roots

  2. Imaginary roots

  3. Exactly one root

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 4a+2b+c=0, so c = -4a-2b. The discriminant D = b^2 - 4ac = b^2 - 4a(-4a-2b) = b^2 + 16a^2 + 8ab. Since ab > 0, 8ab > 0. Also b^2 and 16a^2 are positive. Thus D > 0, implying real roots.

AI explanation

The discriminant of the quadratic equation is b^2-4ac. We know that f(2) = 4a+2b+c = 0, so we can substitute 2b = -4a-c into the discriminant formula. This gives b^2-4ac = ((-4a-c)/2)^2 - 4ac = 4a^2+2ac+(c^2/4)-4ac = 4a^2-2ac+(c^2/4) = (2a-(c/2))^2. Since any real number squared is non-negative, the discriminant is greater than or equal to zero, meaning the equation has real roots.

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