NBT NBT Mathematics: Exponents, Surds and Equations 5 — Questions and Answers
Question 1: Simplify: $\left(\dfrac{x^4}{y^2}\right)^{-1/2}$
- $\dfrac{y}{x^2}$ (Correct answer)
- $\dfrac{x^2}{y}$
- $\dfrac{y^2}{x^4}$
- $\dfrac{x}{y^2}$
Correct answer: $\dfrac{y}{x^2}$
Applying the negative exponent inverts the fraction and the half-power: $\dfrac{y}{x^2}$.
Question 2: Solve for $x$: $2^{2x} - 5 \cdot 2^x + 4 = 0$
- $x = 0$ or $x = 2$ (Correct answer)
- $x = 1$ or $x = 4$
- $x = 0$ or $x = 4$
- $x = 2$ or $x = 4$
Correct 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 $x=2$.
Question 3: What is $\sqrt[3]{-64}$?
- $-4$ (Correct answer)
- $4$
- $-8$
- $8$
Correct answer: $-4$
$(-4)^3 = -64$, so $\sqrt[3]{-64} = -4$.
Question 4: Solve for $x$: $\sqrt{x+1} = x - 1$
- $x = 3$ (Correct answer)
- $x = 0$ or $x = 3$
- $x = 0$
- $x = 2$
Correct 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 $x=3$.
Question 5: Simplify: $2\sqrt{3} \times 3\sqrt{3}$
- $18$ (Correct answer)
- $6\sqrt{3}$
- $6$
- $9\sqrt{3}$
Correct answer: $18$
$2 \times 3 \times \sqrt{3} \times \sqrt{3} = 6 \times 3 = 18$.
Question 6: If $x^2 - x - 12 = 0$, which values satisfy the equation?
- $x = 4$ or $x = -3$ (Correct answer)
- $x = 3$ or $x = -4$
- $x = 6$ or $x = -2$
- $x = 12$ or $x = -1$
Correct answer: $x = 4$ or $x = -3$
$(x-4)(x+3) = 0$, so $x = 4$ or $x = -3$.
Question 7: What is the value of $\left(\dfrac{1}{8}\right)^{-2/3}$?
- $4$ (Correct answer)
- $\dfrac{1}{4}$
- $2$
- $\dfrac{1}{2}$
Correct answer: $4$
$\left(\dfrac{1}{8}\right)^{-2/3} = 8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$.
NBT NBT Mathematics: Exponents, Surds and Equations 5
Simplify: $\left(\dfrac{x^4}{y^2}\right)^{-1/2}$