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

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1. Answer to Question 1

   1. \((x+2)(x+3)=0\), so \(x=-2\) or \(x=-3\) We need two numbers with product 6 and sum 5: 2 and 3. If a product is zero, one of the factors must be zero, so \(x+2=0\) or \(x+3=0\).
   2. \((x-5)(x+3)=0\), so \(x=5\) or \(x=-3\) We need product \(-15\) and sum \(-2\): \(-5\) and 3. Check: \(25-10-15=0\).
   3. \(x(x-9)=0\), so \(x=0\) or \(x=9\) Take out the common factor \(x\). Do not divide both sides by \(x\), because that would lose the solution \(x=0\).
   4. \((x-7)(x+7)=0\), so \(x=7\) or \(x=-7\) This is a difference of two squares, so there are two solutions, one positive and one negative.

2. Answer to Question 2

   1. \(x^2-5x-14=0\), \((x-7)(x+2)=0\), so \(x=7\) or \(x=-2\) Subtract \(5x\) and 14 from both sides so that one side is zero. Check: \(49=35+14\).
   2. \(x^2+x-12=0\), \((x+4)(x-3)=0\), so \(x=-4\) or \(x=3\) Subtract 12 from both sides first. Check: \(9+3=12\).
   3. \(x^2-3x-18=0\), \((x-6)(x+3)=0\), so \(x=6\) or \(x=-3\) Expand the bracket and subtract 18. Check: \(6\times3=18\) and \(-3\times(-6)=18\).
   4. \(x+2=\pm5\), so \(x=3\) or \(x=-7\) Take the square root of both sides, remembering both the positive and the negative root. Check: \(5^2=25\) and \((-5)^2=25\).

3. Answer to Question 3

   1. \(x=-2\): \(4+4-3=5\); \(-1\): \(1+2-3=0\); \(0\): \(-3\); \(1\): \(1-2-3=-4\); \(2\): \(4-4-3=-3\); \(3\): \(9-6-3=0\); \(4\): \(16-8-3=5\). So 5, 0, −3, −4, −3, 0, 5.
   2. The curve crosses the \(x\)-axis where \(y=0\): \(x=-1\) and \(x=3\).
   3. The lowest point is \((1,-4)\).
   4. The line of symmetry passes through the turning point: \(x=1\).

4. Answer to Question 4

   1. \((x-1)(x-5)=0\), so \((1,0)\) and \((5,0)\). The curve meets the \(x\)-axis where \(y=0\). We need product 5 and sum \(-6\): \(-1\) and \(-5\).
   2. When \(x=0\), \(y=\) 5, so \((0,5)\). The \(y\)-intercept is the constant term, because every other term is zero when \(x=0\).
   3. Halfway between 1 and 5: \(x=3\). A parabola is symmetrical, so the line of symmetry is halfway between the two roots.
   4. At \(x=3\): \(y=9-18+5=-4\). The turning point is \((3,-4)\), a minimum (the coefficient of \(x^2\) is positive). Substitute \(x=3\) into the equation to find the \(y\)-coordinate.

5. Answer to Question 5

   1. Roots: \(x(x-4)=0\) gives \((0,0)\) and \((4,0)\). U-shape with minimum at \((2,-4)\).
   2. Roots \(x=\pm2\), \(y\)-intercept 4. An upside-down U (maximum at \((0,4)\)).
   3. Roots \(x=2\) and \(x=-3\), \(y\)-intercept \(-6\). U-shape with a minimum at \(x=-0.5\), \(y=-6.25\).
   4. \(-(x^2-2x-3)=-(x-3)(x+1)\), roots 3 and \(-1\), \(y\)-intercept 3. Upside-down U with maximum at \((1,4)\).

6. Answer to Question 6

   1. The line \(y=2\) crosses the curve at about \(x=-1.4\) and \(x=3.4\) (exact: \(x=1\pm\sqrt6\)).
   2. At \(y=-3\): \(x=0\) and \(x=2\). Read across from \(y=-3\) on the \(y\)-axis to the curve and down to the \(x\)-axis.
   3. At \(y=0\): \(x=-1\) and \(x=3\). These are the roots, where the curve crosses the \(x\)-axis.
   4. The lowest value of \(y\) is \(-4\), so the line \(y=-5\) never meets the graph: the equation has no solutions.

7. Answer to Question 7

   1. \(20\times3-5\times9=60-45=\) 15 m ✓
   2. The turning point is at \(t=2\) (halfway between the zeros): \(h=40-20=\) 20 m.
   3. \(5t(4-t)=0\), so \(t=0\) and \(t=4\): it lands after 4 seconds.
   4. \(5t^2-20t+15=0\), \(t^2-4t+3=0\), \((t-1)(t-3)=0\), so \(t=1\) or \(3\). It is above 15 m for \(3-1=2\) seconds.

8. Answer to Question 8

   1. \(A=x(12-x)=12x-x^2\).
   2. \(x^2-12x+32=0\), \((x-4)(x-8)=0\), so \(x=4\) or \(x=8\) (a \(4\times8\) rectangle, the same shape).
   3. By symmetry the maximum is halfway between 0 and 12, at \(x=6\), area \(=6\times6=\) 36 m².
   4. A square (\(6\text{ m}\times6\text{ m}\)) gives the greatest area for a fixed perimeter.

9. Answer to Question 9

   1. At \(x=0\) and \(x=8\) the height is 0 m.
   2. By symmetry \(x=4\): \(h=\dfrac{4\times4}{4}=\) 4 m.
   3. \(x(8-x)=12\), \(x^2-8x+12=0\), \((x-2)(x-6)=0\): \(x=2\) and \(x=6\).
   4. The arch is at least 3 m high between \(x=2\) and \(x=6\), so the gap is 4 m wide.

10. Answer to Question 10

    1. Area \(=x(x+3)=40\), so \(x^2+3x=40\) and \(x^2+3x-40=0\). The area of a rectangle is width times length.
    2. \((x+8)(x-5)=0\), so \(x=-8\) or \(x=5\). We need two numbers with product \(-40\) and sum 3: 8 and \(-5\).
    3. A length cannot be negative, so \(x=5\) only. So the width of the lawn is 5 m.
    4. Width 5 m, length 8 m, perimeter \(=2(5+8)=\) 26 m. Check: \(5\times8=40\) m², so the perimeter of the lawn is 26 m.

11. Answer to Question 11

    1. \(x^2-10x+16=0\), \((x-2)(x-8)=0\): £2 and £8.
    2. Halfway between 2 and 8: £5, with \(P=-25+50-16=\) £9.
    3. \(-9+30-16=\) £5. Substitute \(x=3\) into the formula. The profit is £5.
    4. An upside-down U. The roots are the prices at which the profit is zero: below £2 or above £8 the stall makes a loss.

12. Answer to Question 12

    1. \(x^2+(x+7)^2=169\), so \(2x^2+14x+49=169\), \(2x^2+14x-120=0\), and dividing by 2: \(x^2+7x-60=0\). The two shorter sides are \(x\) and \(x+7\), and the hypotenuse is 13. Expand \((x+7)^2=x^2+14x+49\).
    2. \((x+12)(x-5)=0\), so \(x=-12\) or \(x=5\). The length must be positive, so \(x=5\). We need product \(-60\) and sum 7: 12 and \(-5\).
    3. 5 m, 12 m, 13 m. The sides are \(x=5\), \(x+7=12\) and the hypotenuse 13.
    4. \(5^2+12^2=25+144=169=13^2\) ✓ By the converse of Pythagoras' theorem, the triangle has a right angle.

13. Answer to Question 13

    1. \(10-5\times1=\) 5 m.
    2. \(10-5t^2=0\), \(t^2=2\), \(t=\sqrt2=\) 1.41 s. Set the height equal to zero, divide by 5 and take the positive square root, because time cannot be negative.
    3. The graph is part of an upside-down parabola with \(h=10\) at \(t=0\), falling to 0 at \(t=1.41\).
    4. \(10-5t^2=5\), \(t^2=1\), so \(t=1\) s (the same as part (a)).

14. Answer to Question 14

    1. \(y=0\) when \(x=0\) or \(x=50\): the ball lands 50 m away.
    2. The maximum is halfway at \(x=25\): \(y=\dfrac{25\times25}{50}=\) 12.5 m.
    3. \(y=\dfrac{10\times40}{50}=\) 8 m.
    4. \(x(50-x)=150\), \(x^2-50x+150=0\), \(x=\dfrac{50\pm\sqrt{2500-600}}{2}=\dfrac{50\pm43.59}{2}\), so \(x=3.2\) m and \(x=46.8\) m. So the ball is 3 m high at about 3.2 m and 46.8 m from the goalkeeper.

15. Answer to Question 15

    1. \(n(n+1)=156\), \(n^2+n-156=0\), \((n-12)(n+13)=0\), so the numbers are 12 and 13.
    2. \(x(x+5)=84\), \(x^2+5x-84=0\), \((x-7)(x+12)=0\). Positive: 7 and 12. (Also \(-12\) and \(-7\).)
    3. \(x^2+x-90=0\), \((x+10)(x-9)=0\), so the positive number is 9.
    4. \(x^2+x+1=0\) has discriminant \(1-4=-3\lt 0\), so there are no real solutions. A negative discriminant means the graph of \(y=x^2+x+1\) never crosses the \(x\)-axis.

16. Answer to Question 16

    1. \(2x^2+8x-3x-12=2x(x+4)-3(x+4)=(2x-3)(x+4)=0\), so \(x=1.5\) or \(x=-4\) Multiply \(a\times c=2\times(-12)=-24\) and find two numbers with product \(-24\) and sum 5: 8 and \(-3\). Check: \((2x-3)(x+4)=2x^2+5x-12\).
    2. \(3x^2-12x+x-4=3x(x-4)+(x-4)=(3x+1)(x-4)=0\), so \(x=-\tfrac13\) or \(x=4\) Here \(ac=-12\), and \(-12\) and 1 have sum \(-11\). Check: \((3x+1)(x-4)=3x^2-11x-4\).
    3. \(6x^2-9x+2x-3=3x(2x-3)+(2x-3)=(3x+1)(2x-3)=0\), so \(x=-\tfrac13\) or \(x=1.5\) Here \(ac=-18\), and \(-9\) and 2 have sum \(-7\). Check: \((3x+1)(2x-3)=6x^2-7x-3\).
    4. \(x\)-axis: \((1.5,0)\) and \((-4,0)\). \(y\)-axis: \(x=0\) gives \(y=-12\), so \((0,-12)\). The \(x\)-intercepts are the solutions from part (a), and the \(y\)-intercept is the constant term.

17. Answer to Question 17

    1. \((x+3)^2-9+1=\) \((x+3)^2-8\) Halve the coefficient of \(x\) to get 3, then subtract \(3^2=9\) to keep the expression the same.
    2. The minimum is when \((x+3)^2=0\): \((-3,-8)\). A square is never negative, so \(y\) is smallest when the bracket is zero, at \(x=-3\).
    3. \((x+3)^2=8\), \(x+3=\pm\sqrt8\), so \(x=-3\pm2\sqrt2\) Rearrange the completed square, take the square root of both sides and simplify \(\sqrt8=2\sqrt2\).
    4. \((x-2)^2-4-3=(x-2)^2-7\). Then \((x-2)^2=7\), \(x=2\pm\sqrt7\), so \(x=4.65\) or \(x=-0.65\). Halve \(-4\) to get \(-2\), then subtract \((-2)^2=4\). Using \(\sqrt7=2.6458\), the solutions are 4.65 and \(-0.65\).

18. Answer to Question 18

    1. \(x=\dfrac{-3\pm\sqrt{9+32}}{4}=\dfrac{-3\pm\sqrt{41}}{4}\), so \(x=0.85\) or \(x=-2.35\) Here \(a=2\), \(b=3\) and \(c=-4\), so \(b^2-4ac=9+32=41\). The solutions are 0.85 and \(-2.35\).
    2. \(x=\dfrac{5\pm\sqrt{25-12}}{2}=\dfrac{5\pm\sqrt{13}}{2}\), so \(x=4.30\) or \(x=0.70\) Here \(a=1\), \(b=-5\) and \(c=3\), so \(-b=5\) and \(b^2-4ac=13\). The solutions are 4.30 and 0.70.
    3. \(b^2-4ac=4-12=-8\lt 0\), so there are no real roots. We cannot take the square root of a negative number, so the graph of \(y=3x^2+2x+1\) does not cross the \(x\)-axis.
    4. Two different real roots need \(16-4k>0\), so \(k\lt 4\). Here \(b^2-4ac=4^2-4\times1\times k\), and the discriminant must be positive for two different roots.

19. Answer to Question 19

    1. Area \(=w(w+5)=120\), so \(w^2+5w-120=0\).
    2. \(w=\dfrac{-5\pm\sqrt{25+480}}{2}=\dfrac{-5\pm\sqrt{505}}{2}\), so \(w=8.74\) or \(w=-13.74\). Here \(a=1\), \(b=5\), \(c=-120\), so \(b^2-4ac=25+480=505\).
    3. The width must be positive: \(w=\) 8.74 m, length \(=\) 13.74 m.
    4. Perimeter \(=2(8.74+13.74)=\) 44.9 m (to 1 d.p., using 44.96).

20. Answer to Question 20

    1. \(b^2-4ac=(-4)^2-4\times1\times k=16-4k\).
    2. One repeated root needs \(16-4k=0\), so \(k=4\): \(x^2-4x+4=(x-2)^2\), touching at \((2,0)\).
    3. No real roots needs \(16-4k\lt 0\), so \(k>4\). Divide both sides by \(-4\) and reverse the inequality sign.
    4. \((x-2)(x-5)=0\): roots 2 and 5. Sum \(=7\) ✓, product \(=10\) ✓.