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4x - 5 - x² = q - (x + p)² where p and q are integers - Edexcel - A-Level Maths Pure - Question 10 - 2012 - Paper 2

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4x---5---x²-=-q---(x-+-p)²--where-p-and-q-are-integers-Edexcel-A-Level Maths Pure-Question 10-2012-Paper 2.png

4x - 5 - x² = q - (x + p)² where p and q are integers. (a) Find the value of p and the value of q. (b) Calculate the discriminant of 4x - 5 - x² (c) On the axes ... show full transcript

Worked Solution & Example Answer:4x - 5 - x² = q - (x + p)² where p and q are integers - Edexcel - A-Level Maths Pure - Question 10 - 2012 - Paper 2

Step 1

Find the value of p and the value of q.

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Answer

To reformat the given equation, we can rearrange:

4x−5−x2=q−(x+p)24x - 5 - x^2 = q - (x + p)^2

Expanding the right side gives:

4x−5−x2=q−(x2+2px+p2)4x - 5 - x^2 = q - (x^2 + 2px + p^2)

Rearranging, we collect the coefficients:

4x−5−x2=q−x2−2px−p2 ⇒(4+2p)x+(p2+q−5)=04x - 5 - x^2 = q - x^2 - 2px - p^2\ \Rightarrow (4 + 2p)x + (p^2 + q - 5) = 0

From this we can determine:

  1. For the coefficients of x:

ightarrow p = 2$$ 2. For the constant terms:

ightarrow 2^2 + q - 5 = 0 ightarrow q = 1$$ Thus, the values are: - $$p = 2$$ - $$q = 1$$

Step 2

Calculate the discriminant of 4x - 5 - x².

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Answer

The given quadratic is:

−x2+4x−5=0-x^2 + 4x - 5 = 0

The discriminant (D) formula is given by:

D=b2−4acD = b^2 - 4ac

Where:

  • a=−1a = -1
  • b=4b = 4
  • c=−5c = -5

Substituting these values gives:

D=42−4(−1)(−5)=16−20=−4D = 4^2 - 4(-1)(-5) = 16 - 20 = -4

Thus, the discriminant is:

  • D=−4D = -4, indicating that there are no real roots.

Step 3

On the axes on page 17, sketch the curve with equation y = 4x - 5 - x² showing clearly the coordinates of any points where the curve crosses the coordinate axes.

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Answer

To sketch the curve:

  1. Find the x-intercepts by setting y = 0:
    • The x-intercepts are found by solving:

0 = 4x - 5 - x^2
\Rightarrow x^2 - 4x + 5 = 0

−Usingthequadraticformula: - Using the quadratic formula:

x = \frac{-b \pm \sqrt{D}}{2a} = \frac{4 \pm \sqrt{-4}}{2}\

−Ascalculatedearlier,D=−4,leadingtocomplexx−intercepts.2.∗∗Findthey−interceptbysettingx=0:∗∗−They−interceptis: - As calculated earlier, D = -4, leading to complex x-intercepts. 2. **Find the y-intercept by setting x = 0:** - The y-intercept is:

y = 4(0) - 5 - (0)^2 = -5\

−Thus,thecurvecrossesthey−axisat(0,−5).3.∗∗Shapeofthecurve:∗∗−Sketchshouldshowadownward−openingparabolawithvertexatamaximumandcrossingthey−axisat(0,−5).Norealrootsindicateitdoesnotcrossthex−axis. - Thus, the curve crosses the y-axis at (0, -5). 3. **Shape of the curve:** - Sketch should show a downward-opening parabola with vertex at a maximum and crossing the y-axis at (0, -5). No real roots indicate it does not cross the x-axis.

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