Interleaved — Phase 2

Maths interleaved practice

printable — key stays hidden on paper

Instructions: Solve all problems. Each requires you to first identify which technique fits — the topics are deliberately mixed. Show all working. Marks are shown in brackets. Use $...$ notation for any typed math. No calculators for factoring/identity work.

1. Solve for $x$ and represent the solution on a number line: $3 - 2x \geq 7$ and $x + 1 > -4$ . [4]

2. Factor completely: $x^3 - 27$ . [3]

3. Find the remainder when $2x^3 - 3x^2 + 4x - 5$ is divided by $(x - 2)$ using the remainder theorem. [3]

4. Solve the simultaneous equations by elimination: $3x + 2y = 16$ and $5x - 2y = 8$ . [4]

5. Simplify and rationalize the denominator: $\dfrac{3}{\sqrt{5} - \sqrt{2}}$ . [4]

6. The sum of two consecutive odd integers is 5 less than three times the smaller. Find the integers. [4]

7. Without solving, state the nature of the roots of $2x^2 - 5x + 4 = 0$ , and find the sum and product of its roots. [4]

8. Solve the radical equation, checking for extraneous solutions: $\sqrt{2x + 3} = x - 6$ . [5]

9. Expand using an appropriate identity: $(2a - 3b)^3$ . [3]

10. Solve the absolute value inequality and write the answer in interval notation: $|2x - 1| < 7$ . [4]

Total: 38 marks

Answer keyMark scheme & solutions

1. (Tests 2.1.11 + 2.1.12 — compound inequality, AND) Why: Two inequalities joined by "and" ⇒ intersection of solution sets.

$3 - 2x \geq 7 \Rightarrow -2x \geq 4 \Rightarrow x \leq -2$ (flip sign, divide by $-2$ ).
$x + 1 > -4 \Rightarrow x > -5$ .
Intersection: $-5 < x \leq -2$ .
Number line: open circle at $-5$ , closed at $-2$ , shade between.

2. (Tests 2.1.6 + 2.1.5 — factoring via $a^3 - b^3$ identity) Why: Difference of cubes, not a common factor or quadratic. $x^3 - 27 = x^3 - 3^3 = (x-3)(x^2 + 3x + 9)$ . The quadratic factor has discriminant $9 - 36 < 0$ , so it's irreducible over reals — fully factored.

3. (Tests 2.1.15 — remainder theorem) Why: Remainder on division by $(x-a)$ is $p(a)$ ; no long division needed. $p(2) = 2(8) - 3(4) + 4(2) - 5 = 16 - 12 + 8 - 5 = 7$ . Remainder $= 7$ .

4. (Tests 2.1.10 — elimination) Why: $+2y$ and $-2y$ cancel directly ⇒ elimination is fastest. Add: $8x = 24 \Rightarrow x = 3$ . Sub: $9 + 2y = 16 \Rightarrow y = \tfrac{7}{2}$ . Solution: $x = 3,\ y = \tfrac{7}{2}$ .

5. (Tests 2.1.22 — rationalization) Why: Surd denominator ⇒ multiply by conjugate. $\dfrac{3}{\sqrt5 - \sqrt2}\cdot\dfrac{\sqrt5 + \sqrt2}{\sqrt5 + \sqrt2} = \dfrac{3(\sqrt5 + \sqrt2)}{5 - 2} = \dfrac{3(\sqrt5+\sqrt2)}{3} = \sqrt5 + \sqrt2$ .

6. (Tests 2.1.8 — word problem → linear equation) Why: Translate to one variable; "consecutive odd" ⇒ $n, n+2$ . $n + (n+2) = 3n - 5 \Rightarrow 2n + 2 = 3n - 5 \Rightarrow n = 7$ . Integers: $7$ and $9$ . (Check: $7+9=16$ ; $3(7)-5=16$ ✓)

7. (Tests 2.1.18 + 2.1.19 — discriminant + Vieta's) Why: "Without solving" signals discriminant/Vieta, not factoring. $D = (-5)^2 - 4(2)(4) = 25 - 32 = -7 < 0$ ⇒ complex (non-real) conjugate roots. Sum $= -b/a = 5/2$ ; Product $= c/a = 4/2 = 2$ .

8. (Tests 2.1.23 — radical equation, extraneous check) Why: Isolate radical, square, then verify (squaring can introduce false roots). $2x + 3 = (x-6)^2 = x^2 - 12x + 36 \Rightarrow x^2 - 14x + 33 = 0$ . $(x-11)(x-3)=0 \Rightarrow x = 11$ or $x = 3$ . Check $x=11$ : $\sqrt{25}=5 = 11-6$ ✓. Check $x=3$ : $\sqrt9 = 3 \neq -3$ ✗ (extraneous). Solution: $x = 11$ .

9. (Tests 2.1.5 — $(a-b)^3$ identity) Why: Cube of binomial ⇒ apply $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ . With $a=2a,\ b=3b$ : $8a^3 - 3(4a^2)(3b) + 3(2a)(9b^2) - 27b^3 = 8a^3 - 36a^2b + 54ab^2 - 27b^3$ .

10. (Tests 2.1.24 — absolute value inequality) Why: $|u| < k$ ⇒ $-k < u < k$ (a "sandwich," not two separate cases). $-7 < 2x - 1 < 7 \Rightarrow -6 < 2x < 8 \Rightarrow -3 < x < 4$ . Interval: $(-3, 4)$ .

[
  {"claim":"Remainder of 2x^3-3x^2+4x-5 div (x-2) is 7","code":"from sympy import symbols, div, Poly\nx=symbols('x')\nq,r=div(2*x**3-3*x**2+4*x-5, x-2, x)\nresult = (r==7)"},
  {"claim":"Radical eqn sqrt(2x+3)=x-6 has only x=11 as valid solution","code":"from sympy import symbols, sqrt, solve\nx=symbols('x')\nsols=solve(sqrt(2*x+3)-(x-6), x)\nresult = (sols==[11])"},
  {"claim":"System 3x+2y=16,5x-2y=8 gives x=3,y=7/2","code":"from sympy import symbols, solve, Rational\nx,y=symbols('x y')\ns=solve([3*x+2*y-16,5*x-2*y-8],[x,y])\nresult = (s[x]==3 and s[y]==Rational(7,2))"}
]