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

Step 1: Write the discriminant.
The equation \((1-p)x^2+4x+p=0\) is a genuine quadratic in \(x\) because \(1-p \neq 0\) for \(0<p<1\). For \(ax^2+bx+c=0\) the discriminant is \(\Delta=b^2-4ac\); here \(a=1-p\), \(b=4\), \(c=p\), so \[ \Delta = 4^2-4(1-p)(pp) = 16-4p+4p^2 \]

Step 2: Show the discriminant is always positive on \(0<p<1\).
Look at \(\Delta=4p^2-4p+16\) as a quadratic in \(p\) itself. Its own discriminant is \((-4)^2-4(44)(16)=16-256=-240\), which is negative, so \(4p^2-4p+16\) never touches zero and keeps one sign for every real \(p\). Since its leading coefficient \(4\) is positive, the expression stays positive everywhere, so in particular \(\Delta>0\) for \(0<p<1\). This means the roots are real and unequal, which already rules out the possibility of imaginary roots.

Step 3: Use the sum and product of the roots to fix the signs.
For \((1-p)x^2+4x+p=0\), the product of roots is \(\dfrac{c}{a}=\dfrac{p}{1-p}\) and the sum of roots is \(-\dfrac{b}{a}=-\dfrac{4}{1-p}\). Since \(0<p<1\), both \(p\) and \(1-p\) are positive numbers, so the product \(\dfrac{p}{1-p}\) is positive. Two roots whose product is positive must carry the same sign, so the roots cannot be of opposite sign; they also cannot be \(0\), since a positive product can never come from a zero root.
Now check the sum: \(1-p>0\) makes \(\dfrac{4}{1-p}\) a positive number, and the minus sign in front makes the sum \(-\dfrac{4}{1-p}\) negative. If both roots were positive their sum would be positive, so a negative sum together with a positive product forces both roots to be negative, and rules out the case of both roots being positive.

Step 4: Check with one value of p.
Take \(p=0.5\): the equation becomes \(0.5x^2+4x+0.5=0\), i.e. \(x^2+8x+1=0\), giving \(x=\dfrac{-8\pm\sqrt{64-4}}{2}=\dfrac{-8\pm\sqrt{60}}{2}\). Since \(\sqrt{60}\approx7.75\) is less than \(8\), both roots come out negative, matching the general result above.

Final Answer:
For every \(p\) with \(0<p<1\), the roots of the equation are real and both negative. \[ \boxed{\text{Real and both negative}} \]