Algebra
Quadratic Equations
MMTS_Full_Test_18
Grade 12
Question:
Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation $x^2+px+\frac{5p}{4}=0$ are rational. Then the area of region $\{(x,y):0\le y\le(x-q)^2, 0\le x\le q\}$ is (in square units)
243
25
$\dfrac{125}{3}$
164
Step-by-Step Solution
Key Concept: Discriminant $p^2-5p=p(p-5)$ must be a perfect square
$p^2-5p=k^2$; for $p=9$: $81-45=36=6^2$ ✓. So $q=9$. Area $=\int_0^9(x-9)^2dx=\left[\frac{(x-9)^3}{3}\right]_0^9=\frac{0-(-729)}{3}=243$.
Correct Answer: 1