Further Mathematics · OCR H245 · Topic 4.05

Further algebra

Roots of equations, transformation of equations, partial fractions.

  • 3 sub-topics
  • 2 in Year 12, 1 in Year 13
  • 47 practice questions
  • 3 animated lessons
Practice questions (47) Lessons Worksheets coming soon All OCR Further Mathematics topics

What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 4.05aRoots of equationsY12
  2. 4.05bTransformation of equationsY12
  3. 4.05cPartial fractionsY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 47 questions in this topic, free, no account needed.

Further algebra and functions

Easy

Roots of a quadratic · Year 12 · 1 mark

  1. The roots of x2+5x−4=0x^2 + 5x - 4 = 0 are α\alpha and β\beta. What are α+β\alpha + \beta and αβ\alpha\beta?
    (1)

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  1. α+β=−ba\alpha + \beta = -\dfrac{b}{a} and αβ=ca\alpha\beta = \dfrac{c}{a}.BB: α+β\alpha + \beta=−5, αβ{}= -5, \ \alpha\beta=−4{}= -4

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Further algebra and functions

Medium

Roots and coefficients · Year 12 · 4 marks

α\alpha, β\beta, γ\gamma are the roots of x3−2x2−8x−7=0x^{3} - 2x^{2} - 8x - 7 = 0.
  1. (a)
    Find α2\alpha^{2}+β2{}+ \beta^{2}+γ2{}+ \gamma^{2}.
    (2)
  2. (b)
    Find 1/α+1/β+1/γ1/\alpha + 1/\beta + 1/\gamma.
    (2)

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Part (a)2 marks

  1. Use the identity.(∑α)2(\sum\alpha)^2−2∑αβ{}- 2\sum\alpha\beta=4−(−16){}= 4 - (-16)
  2. 2020

Part (b)2 marks

  1. ∑αβαβγ\frac{\sum\alpha\beta}{\alpha\beta\gamma}=−87{}= \frac{-8}{7}
  2. −87-\tfrac{8}{7}

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Further algebra and functions

Hard

Transforming roots · Year 12 · 3 marks

α\alpha, β\beta, γ\gamma are the roots of x3−2x2−6x+6=0x^{3} - 2x^{2} - 6x + 6 = 0.
  1. Find a cubic with roots 3α3\alpha, 3β3\beta, 3γ3\gamma.
    (3)
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  1. Substitution method.ww=3x{}= 3x⇒x{}\Rightarrow x=w3{}= \tfrac{w}{3}
  2. (w3)3\left(\tfrac{w}{3}\right)^3−2(w3)2{}- 2\left(\tfrac{w}{3}\right)^2−6w3{}- 6\tfrac{w}{3}+6{}+ 6=0{}= 0
  3. Multiply through by k3k^3.w3−6w2−54w+162=0w^{3} - 6w^{2} - 54w + 162 = 0

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