1. The problem asks to find quadratic equations with integer coefficients given the solutions. 2. Recall that if $x=r_1$ and $x=r_2$ are roots, the quadratic equation is given by: $$a(x-r_1)(x-r_2)=0$$ where $a$ is a nonzero integer. 3. For the first set of roots $x=-5$ and $x=3$: $$a(x+5)(x-3)=0$$ Expanding: $$a(x^2 - 3x + 5x - 15) = a(x^2 + 2x - 15) = 0$$ Choosing $a=1$ for simplicity: $$x^2 + 2x - 15 = 0$$ 4. For the second set of roots $x=\frac{3}{2}$ and $x=-\frac{1}{4}$, to get integer coefficients, multiply the factors by the denominators: $$a\left(x - \frac{3}{2}\right)\left(x + \frac{1}{4}\right) = 0$$ Multiply both factors by 4 to clear denominators: $$a(2x - 3)(4x + 1) = 0$$ Expanding: $$a(8x^2 + 2x - 12x - 3) = a(8x^2 - 10x - 3) = 0$$ Choosing $a=1$: $$8x^2 - 10x - 3 = 0$$ Final answers: - For $x=-5$ and $x=3$: $$x^2 + 2x - 15 = 0$$ - For $x=\frac{3}{2}$ and $x=-\frac{1}{4}$: $$8x^2 - 10x - 3 = 0$$