Edexcel P1 2019 October — Question 8 5 marks

Exam BoardEdexcel
ModuleP1 (Pure Mathematics 1)
Year2019
SessionOctober
Marks5
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Mark schemeDownload PDF ↗
TopicSolving quadratics and applications
TypeQuadratic in x^(1/2) - substitution u = √x
DifficultyStandard +0.3 This is a standard substitution question where students let u = x^(1/2), solve the resulting quadratic 4 - 6u + u² = 0, then back-substitute. The algebraic manipulation and simplification to surd form is routine for P1 level, making it slightly easier than average but not trivial due to the surd simplification requirement.
Spec1.02b Surds: manipulation and rationalising denominators1.02f Solve quadratic equations: including in a function of unknown
8. Solve, using algebra, the equation $$x - 6 x ^ { \frac { 1 } { 2 } } + 4 = 0$$ Fully simplify your answers, writing them in the form \(a + b \sqrt { c }\), where \(a , b\) and \(c\) are integers to be found.
(5)
Question 8:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x - 6x^{\frac{1}{2}} + 4 = 0\); \(x^{\frac{1}{2}} = 3 \pm \sqrt{5}\)M1 A1 M1 for solving \(y^2 - 6y + 4 = 0\) by completing square or quadratic formula; A1 both values, accept \(\frac{6 \pm 2\sqrt{5}}{2}\)
\(x = (3 \pm \sqrt{5})^2 \Rightarrow x = 14 \pm 6\sqrt{5}\)M1 A1 A1 M1 for squaring \(p \pm q\sqrt{r}\) form; A1 for either solution; final A1 both solutions fully simplified
Special Case: If no initial working and \(x^{\frac{1}{2}} = 3 \pm \sqrt{5}\) written as first step, M0A0M1A1A1 possible if correct answers achieved.
Alternative:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x + 4 = 6x^{\frac{1}{2}}\); \((x+4)^2 = 36x\)M1 A1 M1 isolates square root and squares; A1 correct squared expression
\(x^2 - 28x + 16 = 0 \Rightarrow (x-14)^2 = 180 \Rightarrow x = 14 \pm \sqrt{180} \Rightarrow x = 14 \pm 6\sqrt{5}\)M1 A1 A1 M1 expands and solves; both solutions must be fully simplified for final A1A1 
## Question 8:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x - 6x^{\frac{1}{2}} + 4 = 0$; $x^{\frac{1}{2}} = 3 \pm \sqrt{5}$ | M1 A1 | M1 for solving $y^2 - 6y + 4 = 0$ by completing square or quadratic formula; A1 both values, accept $\frac{6 \pm 2\sqrt{5}}{2}$ |
| $x = (3 \pm \sqrt{5})^2 \Rightarrow x = 14 \pm 6\sqrt{5}$ | M1 A1 A1 | M1 for squaring $p \pm q\sqrt{r}$ form; A1 for either solution; final A1 both solutions fully simplified |

**Special Case:** If no initial working and $x^{\frac{1}{2}} = 3 \pm \sqrt{5}$ written as first step, M0A0M1A1A1 possible if correct answers achieved.

**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x + 4 = 6x^{\frac{1}{2}}$; $(x+4)^2 = 36x$ | M1 A1 | M1 isolates square root and squares; A1 correct squared expression |
| $x^2 - 28x + 16 = 0 \Rightarrow (x-14)^2 = 180 \Rightarrow x = 14 \pm \sqrt{180} \Rightarrow x = 14 \pm 6\sqrt{5}$ | M1 A1 A1 | M1 expands and solves; both solutions must be fully simplified for final A1A1 |

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8. Solve, using algebra, the equation

$$x - 6 x ^ { \frac { 1 } { 2 } } + 4 = 0$$

Fully simplify your answers, writing them in the form $a + b \sqrt { c }$, where $a , b$ and $c$ are integers to be found.\\
(5)

\hfill \mbox{\textit{Edexcel P1 2019 Q8 [5]}}
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