NBT Mathematics: Exponents, Surds and Equations Flashcards

Simplify: $\left(\dfrac{x^4}{y^2}\right)^{-1/2}$

Answer: $\dfrac{y}{x^2}$

Applying the negative exponent inverts the fraction and the half-power: $\dfrac{y}{x^2}$.

Solve for $x$: $2^{2x} - 5 \cdot 2^x + 4 = 0$

Answer: $x = 0$ or $x = 2$

Let $k = 2^x$: $k^2 - 5k + 4 = 0 \Rightarrow (k-1)(k-4) = 0$, so $k=1$ or $k=4$, giving $x=0$ or $x=2$.

What is $\sqrt[3]{-64}$?

Answer: $-4$

$(-4)^3 = -64$, so $\sqrt[3]{-64} = -4$.

Solve for $x$: $\sqrt{x+1} = x - 1$

Answer: $x = 3$

Squaring: $x+1=(x-1)^2=x^2-2x+1$; so $x^2-3x=0 \Rightarrow x(x-3)=0$; $x=0$ fails check (LHS=1, RHS=-1), so $x=3$.

Simplify: $2\sqrt{3} \times 3\sqrt{3}$

Answer: $18$

$2 \times 3 \times \sqrt{3} \times \sqrt{3} = 6 \times 3 = 18$.

If $x^2 - x - 12 = 0$, which values satisfy the equation?

Answer: $x = 4$ or $x = -3$

$(x-4)(x+3) = 0$, so $x = 4$ or $x = -3$.

What is the value of $\left(\dfrac{1}{8}\right)^{-2/3}$?

Answer: $4$

$\left(\dfrac{1}{8}\right)^{-2/3} = 8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$.