Moderate -0.8 This is a straightforward modulus equation requiring consideration of cases based on critical values x = 0.5 and x = -3. Students need to square both sides or systematically check regions, but the algebra is simple and the method is standard textbook material with no conceptual surprises.
e.g. squaring both sides to obtain 3 terms on both sides: \((4x^2 - 4x + 1 = x^2 + 6x + 9)\); or consider two linear equations \((2x-1) = \pm(x+3)\)
\(3x^2 - 10x - 8 (= 0)\)
A1 1.1
1 correct solution for A1
Obtain \(4\) and \(-\dfrac{2}{3}\)
A1 1.1 [3]
BC; SC one correct solution from one linear equation B1
## Question 2:
Attempt process for finding both values | M1 1.1a | e.g. squaring both sides to obtain 3 terms on both sides: $(4x^2 - 4x + 1 = x^2 + 6x + 9)$; or consider two linear equations $(2x-1) = \pm(x+3)$
$3x^2 - 10x - 8 (= 0)$ | A1 1.1 | 1 correct solution for A1
Obtain $4$ and $-\dfrac{2}{3}$ | A1 1.1 [3] | **BC**; **SC** one correct solution from one linear equation B1
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