HSA.REI.B.4 math practice. Learn by doing.

Here $a = 1$, $b = 0$, and $c = -9$. Substituting into $b^{2} - 4ac$ gives $0^{2} - 4(1)(-9) = 36\textsf{,}$ so there are $2$ distinct real roots.

Here $a = 1$, $b = -5$, and $c = 6$. Substituting into the quadratic formula gives $x = \dfrac{-(-5) \pm \sqrt{(-5)^{2} - 4(1)(6)}}{2(1)} = \dfrac{5 \pm 1}{2}$, so $x = 2$ and $x = 3$.

The equation is already $(x - 4)^{2} = -16\textsf{.}$ A real number squared cannot be negative, so there are $0$ distinct real solutions.

The quadratic factors as $(2x + 5)(x - 3) = 0\textsf{.}$ By the zero-product property, $2x + 5 = 0$ or $x - 3 = 0\textsf{,}$ so $x = -\dfrac{5}{2}$ or $x = 3\textsf{.}$ The positive solution is $3\textsf{.}$

Here $a = 1$, $b = 2$, and $c = 10$. Substituting into $b^{2} - 4ac$ gives $2^{2} - 4(1)(10) = -36\textsf{,}$ so there are $0$ distinct real roots.

Here $a = 2$, $b = 3$, and $c = -5$. Substituting into $b^{2} - 4ac$ gives $3^{2} - 4(2)(-5) = 9 + 40 = 49$.

Subtract $12$: $x^{2} = -12\textsf{.}$ A real number squared cannot be negative, so there are $0$ distinct real solutions.

The quadratic factors as $(4x + 1)(x - 6) = 0\textsf{.}$ By the zero-product property, $4x + 1 = 0$ or $x - 6 = 0\textsf{,}$ so $x = -\dfrac{1}{4}$ or $x = 6\textsf{.}$ The positive solution is $6\textsf{.}$