Question

If \(0 < p < 1\), then the roots of the equation \((1-p)x^2 + 4x + p = 0\) are:

• A. Both 0
• B. Imaginary
• C. Real and both positive
• D. Real and both negative

Hint:

Look at the sign of the product and the sum of the roots using \(0<p<1\), rather than trying to solve for x directly.

Answer 1

The Correct Option is D

A different way to pin down the sign of the roots is to think of the left side as a graph rather than push through the discriminant-of-a-discriminant trick used elsewhere. Let $f(xx) = (1-p)x^2+4x+p$ for a fixed $p$ with $0<p<1$.

1. Direction the parabola opens: the coefficient of $x^2$ is $1-p$, which is positive since $p<1$. So the graph of $f(xx)$ is a parabola opening upward.
2. Value at $x=0$: $f(00) = p$, which is positive since $0<p<1$. So the graph sits above the x-axis at $x=0$.
3. Where the vertex sits: the vertex of $f(xx)$ is at $x = -\dfrac{b}{2a} = -\dfrac{4}{2(1-p)} = -\dfrac{2}{1-p}$, a negative number because $1-p>0$. So the lowest point of the upward parabola lies to the left of $0$.
4. What this forces about the roots: an upward parabola with its lowest point at a negative x-value, that is still positive at $x=0$, can only cross the x-axis at two points that both lie to the left of $0$. It cannot cross once on each side of $0$, since that would need $f(00)$ to be negative, and it is not.
5. Realness of the roots: separately, $\Delta = 16-4p+4p^2 = 4(p^2-p+4)$, and $p^2-p+4 = \left(p-\frac{1}{2}\right)^2+\frac{15}{4}$ is a square plus a positive number, so it is always positive; hence $\Delta>0$ and the two roots are always real, never imaginary.

Putting the two findings together, the equation always has two real roots, and both sit to the left of $0$, so both are negative.

Let's summarize:

• The parabola opens upward and sits above the axis at $x=0$, with its vertex at a negative x-value.
• That combination is only possible if both roots, when real, are negative, and the discriminant check confirms they are always real.

So for every $p$ with $0<p<1$, the roots are real and both negative.