Quadratic equations and graphs – Year 10 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \((x-5)(x+2)=0\) (1), so \(x=5\) or \(x=-2\) (1) We need two numbers with product \(-10\) and sum \(-3\): \(-5\) and 2. Set each bracket equal to zero.
   2. \(x^2+4x-21=0\) (1), \((x+7)(x-3)=0\), so \(x=-7\) or \(x=3\) (1) Rearrange so that one side is zero, then find two numbers with product \(-21\) and sum 4: 7 and \(-3\). Check: \(9+12=21\).

   Check (a): \(25-15-10=0\) and \(4+6-10=0\), so both roots work.

2. Answer to Question 2 (4 marks)

   1. \(x\)-axis: \((x-5)(x+1)=0\), so \((5,0)\) and \((-1,0)\) (1); \(y\)-axis: \((0,-5)\) (1) The curve meets the \(x\)-axis where \(y=0\). The \(y\)-intercept is the constant term, \(-5\).
   2. Line of symmetry \(x=2\) (1). At \(x=2\): \(y=4-8-5=-9\), so \((2,-9)\) (1) The line of symmetry is halfway between the roots \(-1\) and 5, at \(x=2\). The turning point is the minimum point \((2,-9)\).

   Check (b) by completing the square: \(y=(x-2)^2-9\), giving the same turning point.

3. Answer to Question 3 (4 marks)

   1. \(60-20=\) 40 m (1) Substitute \(t=2\): \(30\times2=60\) and \(5\times2^2=20\). The rocket is 40 m high.
   2. \(30t-5t^2=0\), \(5t(6-t)=0\) (1), so \(t=6\) (1) Set the height equal to zero and factorise. \(t=0\) is the launch, so the rocket lands after 6 seconds.
   3. At \(t=3\): \(90-45=\) 45 m (1) The maximum is halfway between the times when \(h=0\), at \(t=3\). The maximum height is 45 m.

4. Answer to Question 4 (4 marks)

   1. \(x(x+6)=55\) so \(x^2+6x-55=0\) (1) The area of a rectangle is width times length; expand the bracket and subtract 55.
   2. \((x+11)(x-5)=0\) (1), so \(x=5\) (reject \(-11\)) (1) We need product \(-55\) and sum 6: 11 and \(-5\). A width cannot be negative, so \(x=5\).
   3. Width 5, length 11, perimeter \(=2\times16=\) 32 cm (1) The rectangle is 5 cm by 11 cm, so the perimeter is 32 cm.

5. Answer to Question 5 (4 marks)

   1. \(x=\dfrac{2\pm\sqrt{4+48}}{6}=\dfrac{2\pm\sqrt{52}}{6}\) (1), so \(x=1.54\) or \(x=-0.87\) (1) Here \(a=3\), \(b=-2\) and \(c=-4\), so \(b^2-4ac=4+48=52\). The solutions are 1.54 and \(-0.87\).
   2. \((x-4)^2-13\) (1) Halve \(-8\) to get \(-4\), then subtract \(16\): \(3-16=-13\).
   3. The minimum value is −13 (1) A square is never negative, so the smallest value happens when \(x-4=0\). The minimum value is \(-13\).

   Check (a): \(3(1.54)^2-2(1.54)-4\approx0\). The exact answers are \(\tfrac{1\pm\sqrt{13}}{3}\).

Total : 20 marks