STAAR Algebra I Quadratic Functions and Equations. Practice it free.

Subtract the fourth-year salary from the first: $\text{\char36}54{,}600 - \text{\char36}48{,}000 = \text{\char36}6{,}600$. Over four years there are three equal raises, so divide by $4 - 1 = 3$ to get $x = \text{\char36}2{,}200$. Add $\text{\char36}2{,}200$ each year after the first: $\text{\char36}48{,}000$, $\text{\char36}50{,}200$, $\text{\char36}52{,}400$, $\text{\char36}54{,}600$. Sum all four years: $\text{\char36}205{,}200$.

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

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

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.

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

Set $y = 0$: $-(x + 8)(x - 2) = 0$. The factor $-1$ is nonzero, so the zeros come from $x + 8 = 0 \Rightarrow x = -8$ and $x - 2 = 0 \Rightarrow x = 2$. Of the choices, $-8$ is an x-coordinate of an x-intercept.

$x^2 - 81 = (x - 9)(x + 9) = 0$, so $x = 9$ or $x = -9$. The positive root is $9$.