Standard +0.3 This is a quadratic-in-disguise problem requiring substitution u = x², solving 4u² + 3u - 1 = 0, then finding x from u values. While it involves multiple steps (substitution, quadratic formula/factorization, taking square roots, rejecting negative u), it's a standard C1 technique with no conceptual surprises, making it slightly easier than average.
Substitute for \(x^2\). No marks if whole equation square rooted etc. No marks if straight to formula with no evidence of substitution at start and no square rooting/squaring at end. If factorising into two brackets: \((4x-1)(x+1) = 0\) M1 A1; \((2x+1)(2x-1)(x+1) = 0\) M1 A1 (BOD). Spotted solutions: If M0 DM0 or M1 DM0. SR B1 \(x = \frac{1}{2}\) www; SC B1 \(x = -\frac{1}{2}\) www. (Can then get 5/5 if both found www and exactly two solutions justified)
$k = x^2$; $4k^2 + 3k - 1 = 0$; $(4k-1)(k+1) = 0$; $k = \frac{1}{4}, k = -1$; $x = \pm\sqrt{\frac{1}{4}}$; $x = \pm\frac{1}{2}$ | M1*, M1dep*, A1, M1, A1, A1 [5] | Substitute for $x^2$. No marks if whole equation square rooted etc. No marks if straight to formula with no evidence of substitution at start and no square rooting/squaring at end. If factorising into two brackets: $(4x-1)(x+1) = 0$ M1 A1; $(2x+1)(2x-1)(x+1) = 0$ M1 A1 (BOD). Spotted solutions: If M0 DM0 or M1 DM0. SR B1 $x = \frac{1}{2}$ www; SC B1 $x = -\frac{1}{2}$ www. (Can then get 5/5 if both found www and exactly two solutions justified)