2026 QUESTION 1 1 Solve for π‘₯: 1.1 2π‘₯ 2 βˆ’ 7π‘₯ = 0 (3) 1.2 π‘₯ 2 = 12π‘₯ βˆ’ 21 (correct to ONE decimal place) (4) 1.3 4π‘₯Β² βˆ’ 12π‘₯ + 8 β‰₯ 0 (4) 1.4 √5 βˆ’ 2π‘₯ βˆ’ π‘₯ 2 = 4 (6) 1.5 2 2π‘₯+1 + 5 . 2 π‘₯ βˆ’ 3 = 0 (5) 2 Given: A = 4 π‘₯ 2+2 ; π‘₯ ∈ ℝ 2.1 Determine the maximum value of A. Explain your answer. (2) 2.2 For which value of π‘₯ will A have a maximum value? (1) [25] QUESTION 2 1 Solve for π‘₯ and 𝑦 simultaneously: π‘₯ βˆ’ 2𝑦 = 2 and π‘₯Β² + π‘₯𝑦 βˆ’ 5π‘₯ + 3𝑦 = 3 (6) 2 Determine integers π‘Ž and 𝑏 so that (3 βˆ’ √2) 2 = π‘Ž + 𝑏 √2 . (3) 3 Anne and Zandi decide to paint a room. Anne paints (4 + 2π‘₯ 2 π‘š2 . Zandi starts later and paints 4 π‘š2 . They continue painting the next day and they both paint a further (2π‘₯) π‘š2 . Determine the total area (in π‘š2 that Anne painted if she painted exactly double the total area that Zandi painted.d (4)

Mathematics

ScanSolve AI Answer

Step-by-step solution

2 steps
  1. Factor out the common term.

    The common factor in the equation 2x2βˆ’7x=02x^2 - 7x = 0 is xx.

    x(2xβˆ’7)=0x(2x - 7) = 0
  2. Set each factor equal to zero and solve for xx.

    This gives two possible solutions:

    x=0x = 0

    or

    2xβˆ’7=02x - 7 = 0 2x=72x = 7 x=72x = \frac{7}{2}

    The solutions are x=0x=0 or x=72x=\frac{7}{2}.

    x=0Β orΒ x=72\boxed{x=0 \text{ or } x=\frac{7}{2}}
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