Moderate -0.8 This is a straightforward modulus equation requiring students to split into two cases (3x+4a = ±5a) and solve two linear equations. It's simpler than average A-level work as it involves only basic algebraic manipulation with no additional complications, though the presence of a parameter 'a' adds minimal complexity.
with signs of \(3x\) and \(5a\) different; allow M1 only if \(a\) given particular value and no recovery occurs; allow M1 only if \(a\) in terms of \(x\) attempted; allow M1 only if double inequality attempted but with no recovery to state actual values of \(x\)
Obtain \(-3a\)
A1
3 as final answer
Or: Obtain \(9x^2 + 24ax + 16a^2 = 25a^2\)
B1
as far as substitution into correct quadratic formula or correct factorisation of their quadratic; allow M1 only if \(a\) given particular value
Attempt solution of 3-term quad eqn
M1
Obtain \(-3a\) and \(\frac{1}{4}a\)
A1 (3)
or equivs; as final answers; and no others
Either: Obtain $\frac{1}{4}a$ | B1 | condone $|x| = \frac{1}{4}a$
Attempt solution of linear eqn | M1 | with signs of $3x$ and $5a$ different; allow M1 only if $a$ given particular value and no recovery occurs; allow M1 only if $a$ in terms of $x$ attempted; allow M1 only if double inequality attempted but with no recovery to state actual values of $x$
Obtain $-3a$ | A1 | 3 as final answer
Or: Obtain $9x^2 + 24ax + 16a^2 = 25a^2$ | B1 | as far as substitution into correct quadratic formula or correct factorisation of their quadratic; allow M1 only if $a$ given particular value
Attempt solution of 3-term quad eqn | M1 |
Obtain $-3a$ and $\frac{1}{4}a$ | A1 (3) | or equivs; as final answers; and no others
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