Practice problems

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Q. Find the sum of all non-negative integers $a$ such that the quadratic equation

$$a^2 x^2 - (3a^2 - 8a)x + 2a^2 - 13a + 15$$

has at least one integer root.

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Solution:

Solution sketch (step-by-step)

Step	Reasoning
1 . Write the equation	$a^{2}x^{2}-\bigl(3a^{2}-8a\bigr)x+\bigl(2a^{2}-13a+15\bigr)=0$.
2 . Compute the discriminant $D$	$D=\bigl(3a^{2}-8a\bigr)^{2}-4a^{2}\bigl(2a^{2}-13a+15\bigr)$.
3 . Simplify $D$	$D=a^{4}+4a^{3}+4a^{2}=a^{2}(a+2)^{2}=\bigl[a(a+2)\bigr]^{2}$ – already a perfect square, so roots are always rational.
4 . Write the roots explicitly	$x=\dfrac{3a^{2}-8a\;\pm\;a(a+2)}{2a^{2}}$. Splitting the ± gives $\displaystyle x_{1}=2-\frac{3}{a},\qquad x_{2}=1-\frac{5}{a}$.
5 . Demand an integer root	For at least one of $x_{1},x_{2}$ to be an integer, its fractional part must vanish: * $x_{1}$ is integer ⇒ $\tfrac{3}{a}$ is integer ⇒ $a\mid 3$ ⇒ $a=1$ or $3$.* $x_{2}$ is integer ⇒ $\tfrac{5}{a}$ is integer ⇒ $a\mid 5$ ⇒ $a=1$ or $5$.
6 . Collect admissible $a$ (non-negative, and $a\neq0$ so the equation stays quadratic)	$a\in\{1,3,5\}$.
7 . Check each $a$ quickly	$a=1:\;x^{2}+5x+4=0$ ⇒ $-1,-4$ OK.$a=3:\;9x^{2}-3x-6=0$ ⇒ $1,\,-\frac23$ OK.$a=5:\;25x^{2}-35x=0$ ⇒ $0,\,\frac74$ ⇒ at least one integer root (0) OK.
8 . Sum them	$1+3+5=9$.

Answer

$$\boxed{9}$$

So the sum of all non-negative integers $a$ for which the given quadratic has at least one integer root is 9.

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