PSAT 8/9 Advanced Math. Practice it free.

Here $a = 1$, $b = 0$, and $c = -25$. Substituting into $b^{2} - 4ac$ gives $0^{2} - 4(1)(-25) = 100\textsf{,}$ so there are $2$ distinct real roots.

Here $a = 1$, $b = 6$, and $c = 9$. Substituting into $b^{2} - 4ac$ gives $6^{2} - 4(1)(9) = 0$, so there is exactly $1$ distinct real solution.

Subtract $15$: $3(x + 2)^{2} = -15\textsf{.}$ Divide by $3$: $(x + 2)^{2} = -5\textsf{.}$ 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 $3\textsf{.}$

The constants $3$ and $8$ are like terms: $3 + 8 = 11$. The term $w$ stays the same, so the expression is equivalent to $w + 11\textsf{.}$

The greatest common factor of $5x^{2}$ and $10x$ is $5x\textsf{.}$ Factoring gives $5x(x + 2)$.

$81 - x^{2} = 9^{2} - x^{2} = (9 - x)(9 + x)$.

Write $24y$ with denominator $c$: $\dfrac{12y}{c} + \dfrac{24yc}{c} = \dfrac{12y + 24yc}{c} = \dfrac{12y(1 + 2c)}{c}$.