Standard +0.3 Part (a) is straightforward factoring (factor out 10x, then solve a quadratic). Part (b) requires recognizing the substitution x = (y+3)^(1/2), then back-substituting to find y values. This is a standard 'hence' question testing algebraic manipulation and substitution, slightly above average due to the fractional powers but still routine for P1.
Attempting to factorise; cancelling/factorising out \(x\) to achieve \(Dx(Ax^2+Bx+C)\) where \(A,B,C \neq 0\), \(D=1,2,5\) or \(10\). If only resulting quadratic \((2x+1)(x-3)\) shown, must state \(x=0\) somewhere
Two of \(x = 0, -\frac{1}{2}, 3\)
A1
Following M1 awarded
All of \(x = 0, -\frac{1}{2}, 3\)
A1
Withhold if any additional solutions. Do not need to be on one line (3 marks)
Part (b):
Answer
Marks
Guidance
Working/Answer
Mark
Guidance
Sets or implies \((y+3)^{\frac{1}{2}} = 0\) or \(-\frac{1}{2}\) or \(3\)
B1ft
Follow through on any of their solutions from (a); \((y+3)=0 \Rightarrow y=-3\) is insufficient for B mark but acceptable for final A mark
Full method to find \(y\)
M1
Full method to find value of \(y\) from non-zero solution for \(x\); scored for squaring and subtracting 3; allow for squaring and subtracting 3 from a negative value
\(y = 6\)
A1ft
Follow through on their positive solution from part (a)
\(y = -3, 6\)
A1
Both values with no incorrect working. If \(-\frac{11}{4}\) is found it must be discounted. Note: \((y+3)^{\frac{1}{2}}=0 \Rightarrow y+3=\sqrt{0} \Rightarrow y=-3\) is incorrect method so A0 (4 marks)
## Question 5:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $20x^3 - 50x^2 - 30x = 0 \Rightarrow 10x(2x^2 - 5x - 3) = 0$ | | |
| $\Rightarrow 10x(2x+1)(x-3) = 0$ | M1 | Attempting to factorise; cancelling/factorising out $x$ to achieve $Dx(Ax^2+Bx+C)$ where $A,B,C \neq 0$, $D=1,2,5$ or $10$. If only resulting quadratic $(2x+1)(x-3)$ shown, must state $x=0$ somewhere |
| Two of $x = 0, -\frac{1}{2}, 3$ | A1 | Following M1 awarded |
| All of $x = 0, -\frac{1}{2}, 3$ | A1 | Withhold if any additional solutions. Do not need to be on one line **(3 marks)** |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Sets or implies $(y+3)^{\frac{1}{2}} = 0$ or $-\frac{1}{2}$ or $3$ | B1ft | Follow through on any of their solutions from (a); $(y+3)=0 \Rightarrow y=-3$ is insufficient for B mark but acceptable for final A mark |
| Full method to find $y$ | M1 | Full method to find value of $y$ from non-zero solution for $x$; scored for **squaring** and **subtracting 3**; allow for squaring and subtracting 3 from a negative value |
| $y = 6$ | A1ft | Follow through on their positive solution from part (a) |
| $y = -3, 6$ | A1 | Both values with no incorrect working. If $-\frac{11}{4}$ is found it must be discounted. Note: $(y+3)^{\frac{1}{2}}=0 \Rightarrow y+3=\sqrt{0} \Rightarrow y=-3$ is incorrect method so A0 **(4 marks)** |
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