Standard +0.3 This is a standard quadratic-by-substitution question (let u = x^(1/2)) requiring routine algebraic manipulation and use of the quadratic formula, with the additional step of squaring to return to x. The surd form answer adds minor complexity but this is a well-practiced C1 technique, making it slightly easier than average.
7 Solve the equation \(x - 6 x ^ { \frac { 1 } { 2 } } + 2 = 0\), giving your answers in the form \(p \pm q \sqrt { r }\), where \(p , q\) and \(r\) are integers.
Use a substitution to obtain a quadratic with \(k^2\), \(6k\) and \(2\) (may be implied by squaring or rooting later). Any sight of 4 or 36x from "squaring" original equation scores 0/6. Alternative solution: \(6\sqrt{x} = x + 2\); \(36x = x^2 + 4x + 4\); Rearrange and square both sides M1*; Correct simplified quadratic \(x^2 - 32x + 4 = 0\) A1; Method to solve quadratic M1dep; Correct unsimplified expression A1; Correct discriminant A1; \(16 \pm 6\sqrt{7}\) o.e. A1
\(k^2 - 6k + 2 = 0\)
\((k-3)^2 - 7 = 0\)
M1 dep
Correct method to solve resulting quadratic
\(k = 3 \pm \sqrt{7}\)
A1
\(k = 3 \pm \sqrt{7}\) or \(k = \frac{6 \pm \sqrt{28}}{2}\) or \(k = 3 \pm \frac{\sqrt{28}}{2}\)
\(x = \left(3 \pm \sqrt{7}\right)^2\)
M1
Recognise the need to square to obtain \(x\). SC If no evidence of substitution at start and no squaring/rooting at end: Correct method for solving quadratic with \(a=1\), \(b=-6\), \(c=2\) and solution simplified to \(3 \pm \sqrt{7}\) B1
M1
Correct method for squaring \(a + \sqrt{b}\) (3 or 4 term expansion)
\(x = 16 + 6\sqrt{7}\) or \(x = 16 - 6\sqrt{7}\)
A1 [6]
Allow \(16 \pm 3\sqrt{28}\) or \(16 \pm 2\sqrt{63}\)
## Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k = x^{\frac{1}{2}}$ | M1* | Use a substitution to obtain a quadratic with $k^2$, $6k$ and $2$ (may be implied by squaring or rooting later). **Any sight of 4 or 36x from "squaring" original equation scores 0/6.** Alternative solution: $6\sqrt{x} = x + 2$; $36x = x^2 + 4x + 4$; Rearrange and square both sides **M1***; Correct simplified quadratic $x^2 - 32x + 4 = 0$ **A1**; Method to solve quadratic **M1dep**; Correct unsimplified expression **A1**; Correct discriminant **A1**; $16 \pm 6\sqrt{7}$ o.e. **A1** |
| $k^2 - 6k + 2 = 0$ | | |
| $(k-3)^2 - 7 = 0$ | M1 dep | Correct method to solve resulting quadratic |
| $k = 3 \pm \sqrt{7}$ | A1 | $k = 3 \pm \sqrt{7}$ or $k = \frac{6 \pm \sqrt{28}}{2}$ or $k = 3 \pm \frac{\sqrt{28}}{2}$ |
| $x = \left(3 \pm \sqrt{7}\right)^2$ | M1 | Recognise the need to square to obtain $x$. SC If no evidence of substitution at start and no squaring/rooting at end: Correct method for solving quadratic with $a=1$, $b=-6$, $c=2$ and solution simplified to $3 \pm \sqrt{7}$ **B1** |
| | M1 | Correct method for squaring $a + \sqrt{b}$ (3 or 4 term expansion) |
| $x = 16 + 6\sqrt{7}$ or $x = 16 - 6\sqrt{7}$ | A1 [6] | Allow $16 \pm 3\sqrt{28}$ or $16 \pm 2\sqrt{63}$ |
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7 Solve the equation $x - 6 x ^ { \frac { 1 } { 2 } } + 2 = 0$, giving your answers in the form $p \pm q \sqrt { r }$, where $p , q$ and $r$ are integers.
\hfill \mbox{\textit{OCR C1 2012 Q7 [6]}}