OCR FP1 2010 June — Question 7 7 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with nonlinearly transformed roots
DifficultyStandard +0.8 This is a roots transformation problem requiring systematic use of Vieta's formulas and algebraic manipulation. While the technique is standard for FP1, students must correctly identify α+β=-2k and αβ=k, compute the new roots as -2k/α and -2k/β, then find their sum and product to construct the new equation. The multi-step algebraic reasoning and careful manipulation elevate this above routine questions, though it follows a well-established method taught in Further Pure.
Spec4.05a Roots and coefficients: symmetric functions

The quadratic equation \(x^2 + 2kx + k = 0\), where \(k\) is a non-zero constant, has roots \(\alpha\) and \(\beta\). Find a quadratic equation with roots \(\frac{\alpha + \beta}{\alpha}\) and \(\frac{\alpha + \beta}{\beta}\). [7]

Either
AnswerMarks Guidance
\(\alpha + \beta = -2k\)B1B1 State or use correct results
\(\alpha\beta = k\)M1 Attempt to find sum of new roots
A1Obtain \(4k\)
M1Attempt to find product of new roots
A1Obtain \(4k\)
B1ft 7Correct quadratic equation a.e.f.
\(y^2 - 4ky + 4k = 0\)
Or
AnswerMarks Guidance
\(\alpha + \beta = -2k\)B1 State or use correct result
\(\alpha\beta = k\)B1 State or imply form of new roots
\(\beta = \frac{-2k}{a}\)B1 State correct substitution
\(y = \frac{-2k}{x}\)M1 Rearrange and substitute for \(x\)
A1Correct unsimplified equation
M1Attempt to clear fractions
A1Correct quadratic equation a.e.f.
\(y^2 - 4ky + 4k = 0\)
Or
AnswerMarks Guidance
\(-k \pm \sqrt{k^2 - k}\)B1 Find roots of original equation
\(\frac{\alpha + \beta}{\alpha} = \frac{2k}{k + \sqrt{k^2 - k}}\)B1 Express both new roots in terms of \(k\)
\(\frac{\alpha + \beta}{\beta} = \frac{2k}{k - \sqrt{k^2 - k}}\)
M1Attempt to find sum of new roots
A1Obtain \(4k\)
M1Attempt to find product of new roots
A1Obtain \(4k\)
B1ft 7Correct quadratic equation a.e.f.
\(y^2 - 4ky + 4k = 0\)
**Either**

$\alpha + \beta = -2k$ | B1B1 | State or use correct results
$\alpha\beta = k$ | M1 | Attempt to find sum of new roots
| A1 | Obtain $4k$
| M1 | Attempt to find product of new roots
| A1 | Obtain $4k$
| B1ft 7 | Correct quadratic equation a.e.f.

$y^2 - 4ky + 4k = 0$

**Or**

$\alpha + \beta = -2k$ | B1 | State or use correct result
$\alpha\beta = k$ | B1 | State or imply form of new roots
$\beta = \frac{-2k}{a}$ | B1 | State correct substitution
$y = \frac{-2k}{x}$ | M1 | Rearrange and substitute for $x$
| A1 | Correct unsimplified equation
| M1 | Attempt to clear fractions
| A1 | Correct quadratic equation a.e.f.

$y^2 - 4ky + 4k = 0$

**Or**

$-k \pm \sqrt{k^2 - k}$ | B1 | Find roots of original equation
$\frac{\alpha + \beta}{\alpha} = \frac{2k}{k + \sqrt{k^2 - k}}$ | B1 | Express both new roots in terms of $k$
$\frac{\alpha + \beta}{\beta} = \frac{2k}{k - \sqrt{k^2 - k}}$ | | 
| M1 | Attempt to find sum of new roots
| A1 | Obtain $4k$
| M1 | Attempt to find product of new roots
| A1 | Obtain $4k$
| B1ft 7 | Correct quadratic equation a.e.f.

$y^2 - 4ky + 4k = 0$

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The quadratic equation $x^2 + 2kx + k = 0$, where $k$ is a non-zero constant, has roots $\alpha$ and $\beta$. Find a quadratic equation with roots $\frac{\alpha + \beta}{\alpha}$ and $\frac{\alpha + \beta}{\beta}$. [7]

\hfill \mbox{\textit{OCR FP1 2010 Q7 [7]}}