IBDP Maths AHL 2.12 Polynomial functions and their graphs AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
By the remainder theorem, the remainder when \(P(x)\) is divided by \((x+1)\) is \(P(-1)\).
\(P(-1)=12\)
\(2(-1)^2+q(-1)+r=12\)
\(2-q+r=12\)
\(r-q=10\)
Therefore:
\(r=q+10\)
For the quadratic equation \(2x^2+qx+r=0\), the sum and product of the roots are:
\(\alpha+\beta=-\dfrac{q}{2}\)
\(\alpha\beta=\dfrac{r}{2}\)
Use the identity:
\((\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta\)
Substitute the given information.
\(\left(-\dfrac{q}{2}\right)^2=\dfrac{37}{4}+2\left(\dfrac{r}{2}\right)\)
\(\dfrac{q^2}{4}=\dfrac{37}{4}+r\)
\(q^2-4r=37\)
Now substitute \(r=q+10\).
\(q^2-4(q+10)=37\)
\(q^2-4q-77=0\)
\((q-11)(q+7)=0\)
Therefore:
\(q=11\) or \(q=-7\)
If \(q=11\), then \(r=21\). The discriminant of \(2x^2+11x+21\) is:
\(11^2-4(2)(21)=-47\)
This is negative, so the roots would not be real. Therefore, this solution must be rejected because \(\alpha,\beta\in\mathbb{R}\).
Hence:
\(q=-7\)
\(r=q+10=-7+10=3\)
For these values, the discriminant is \((-7)^2-4(2)(3)=25>0\), confirming that the roots are real.
✅ Answer: \(q=-7\) and \(r=3\)