LEAP 2025 Algebra I Expressing Mathematical Reasoning. 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 $3\textsf{.}$

$5^2 = 25 < 27 < 36 = 6^2$, so $\sqrt{27}$ is between $5$ and $6$; the larger is $6$.

The amount after $t$ years is $A(t) = P(1 + r)^{t}\textsf{.}$ Here $P = 500\textsf{,}$ $r = 0.20\textsf{,}$ and $t = 3\textsf{,}$ so $A(3) = 500(1.20)^{3} = 864\textsf{.}$

When $w = 2$, the exponent is $2/2 = 1$. So $L(2) = 180(0.88)^1 = 158.4$ lumens.

$10 \div 5 = 2$ half-lives. Remaining: $\left(\dfrac{1}{2}\right)^{2} = \dfrac{1}{4}$.