Indices, surds and quadratic functions – Year 12 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. The denominator 3 of the power means a cube root and the numerator 2 means squaring; taking the root first keeps the numbers small. \(27^{2/3}=(\sqrt[3]{27})^2=3^2=\) 9. So \(27^{2/3}=9\).
   2. A negative power means the reciprocal; then take the fourth root and cube it. \(16^{-3/4}=\dfrac1{16^{3/4}}=\dfrac1{(\sqrt[4]{16})^3}=\dfrac1{2^3}=\) \(\dfrac18\). So \(16^{-3/4}=\tfrac18\).
   3. A negative power inverts the fraction: \(\left(\dfrac{27}{8}\right)^{2/3}=\left(\dfrac32\right)^2=\) \(\dfrac94\). Check: \(\left(\tfrac94\right)^{-1}=\tfrac49\) and \(\left(\tfrac8{27}\right)^{2/3}=\tfrac49\) ✓.
   4. Use \((ab)^n=a^nb^n\), \(x^m\times x^n=x^{m+n}\) and \(x^m\div x^n=x^{m-n}\). Numerator: \((2x^3)^2\times3x^{-5}=4x^6\times3x^{-5}=12x\). Then \(\dfrac{12x}{6x^{-1}}=2x^{1-(-1)}=\) \(2x^2\). So the expression simplifies to \(2x^2\).

2. Answer to Question 2

   1. Simplify each surd by taking out the largest square factor: \(\sqrt{75}=\sqrt{25\times3}\), \(\sqrt{12}=\sqrt{4\times3}\), \(\sqrt{27}=\sqrt{9\times3}\). \(5\sqrt3+2\sqrt3-3\sqrt3=\) \(4\sqrt3\). Like surds can be collected just like like terms.
   2. Expand the brackets term by term, using \(\sqrt5\times\sqrt5=5\). \(6-3\sqrt5+2\sqrt5-5=\) \(1-\sqrt5\). This is in the form \(a+b\sqrt c\) with \(a=1\), \(b=-1\), \(c=5\).
   3. To rationalise the denominator, multiply by \(\dfrac{\sqrt3}{\sqrt3}\), which equals 1. Multiply top and bottom by \(\sqrt3\): \(\dfrac{6\sqrt3}{3}=\) \(2\sqrt3\). The denominator is now an integer.
   4. For a denominator \(a-\sqrt b\), multiply top and bottom by the conjugate \(a+\sqrt b\); the denominator becomes \(a^2-b\). Multiply by the conjugate \(3+\sqrt5\): \(\dfrac{4(3+\sqrt5)}{9-5}=\dfrac{4(3+\sqrt5)}4=\) \(3+\sqrt5\). So \(\dfrac4{3-\sqrt5}=3+\sqrt5\).
   5. Multiply top and bottom by the conjugate of the denominator. Multiply by \(\sqrt2+1\): \(\dfrac{(\sqrt2+1)^2}{2-1}=2+2\sqrt2+1=\) \(3+2\sqrt2\). Check: \(\dfrac{2.414}{0.414}\approx5.83\) and \(3+2\sqrt2\approx5.83\) ✓.

3. Answer to Question 3

   1. Halve the coefficient of \(x\) to get \((x-4)^2\), then subtract \(4^2=16\) to compensate. \(x^2-8x+19=(x-4)^2-16+19=(x-4)^2+3\). Turning point \((4,3)\), a minimum (positive \(x^2\) coefficient).
   2. Take out the factor \(-2\) from the terms in \(x\) first, then complete the square inside the bracket. \(-2(x^2-6x)-7=-2[(x-3)^2-9]-7=-2(x-3)^2+18-7=-2(x-3)^2+11\). Turning point \((3,11)\), a maximum. A negative coefficient of \(x^2\) gives a maximum.
   3. Use the completed-square form and rearrange to isolate \((x-3)^2\). \(-2(x-3)^2+11=0\) gives \((x-3)^2=\dfrac{11}2\), so \(x=3\pm\sqrt{\dfrac{11}{2}}=3\pm\dfrac{\sqrt{22}}2\). (Check: \(\sqrt{5.5}\approx2.345\).) So the two exact roots are \(x=3\pm\dfrac{\sqrt{22}}2\), approximately 0.655 and 5.35.

4. Answer to Question 4

   1. Look for two numbers with product \(2\times(-12)=-24\) and sum \(-5\): these are \(-8\) and 3. Factorise: \((2x+3)(x-4)=0\), so \(x=\) \(-\tfrac32\) or 4. Check: \(2(16)-20-12=0\) ✓.
   2. The expression does not factorise, so complete the square. Complete the square: \((x+3)^2-9-2=0\), \((x+3)^2=11\), \(x=\) \(-3\pm\sqrt{11}\). These are the exact roots.
   3. Use the quadratic formula with \(a=3\), \(b=-2\), \(c=-4\), and simplify \(\sqrt{52}=2\sqrt{13}\). Formula: \(x=\dfrac{2\pm\sqrt{4+48}}{6}=\dfrac{2\pm\sqrt{52}}6=\dfrac{2\pm2\sqrt{13}}6=\) \(\dfrac{1\pm\sqrt{13}}3\). These are the exact roots, approximately 1.54 and \(-0.869\).

5. Answer to Question 5

   1. Equal roots means the discriminant is zero; here \(a=1\), \(b=k\), \(c=9\). \(k^2-36=0\), so \(k=\) \(\pm6\). Check: \(k=6\) gives \((x+3)^2=0\) ✓.
   2. No real roots means the discriminant is negative; here \(a=2\), \(b=-4\), \(c=k\). \(16-8k<0\), so \(8k>16\) and \(k>2\). So the equation has no real roots when \(k>2\).
   3. Real roots need \(b^2-4ac\ge0\); here \(a=k\), \(b=8\), \(c=2\). \(64-8k\ge0\) gives \(k\le8\). Also \(k\ne0\), otherwise the equation is not quadratic. So \(k\le8\), \(k\ne0\) (two distinct real roots need \(k<8\)).

6. Answer to Question 6

   1. \(x^4=(x^2)^2\), so the equation is a quadratic in \(x^2\). Let \(y=x^2\): \(y^2-13y+36=0\), \((y-4)(y-9)=0\), so \(x^2=4\) or \(9\) and \(x=\) \(\pm2\), \(\pm3\). All four values are solutions because both 4 and 9 are positive.
   2. \(x=(\sqrt x)^2\), so the equation is a quadratic in \(\sqrt x\). Let \(u=\sqrt x\): \(u^2-5u+6=0\), \((u-2)(u-3)=0\), so \(\sqrt x=2\) or 3 and \(x=\) 4 or 9. Check: \(4-10+6=0\) ✓ and \(9-15+6=0\) ✓.
   3. \(2^{2x}=(2^x)^2\), so the equation is a quadratic in \(2^x\). Let \(y=2^x\): \(y^2-6y+8=0\), \((y-2)(y-4)=0\), so \(2^x=2\) or \(4\) and \(x=\) 1 or 2. Check: \(16-24+8=0\) ✓.

7. Answer to Question 7

   1. Multiply length by width; the brackets form a difference of two squares, \((a+b)(a-b)=a^2-b^2\). Area \(=(5+\sqrt2)(5-\sqrt2)=25-2=\) 23 m\(^2\) (difference of two squares). The surds cancel, so the area is a whole number.
   2. The perimeter is twice the sum of the length and the width. Perimeter \(=2[(5+\sqrt2)+(5-\sqrt2)]=2\times10=\) 20 m. The \(\sqrt2\) terms cancel in the sum.
   3. Use Pythagoras with the length and width, expanding each square with \((a\pm b)^2=a^2\pm2ab+b^2\). Diagonal\(^2=(5+\sqrt2)^2+(5-\sqrt2)^2=(27+10\sqrt2)+(27-10\sqrt2)=54\), so the diagonal is \(\sqrt{54}=\sqrt{9\times6}=3\sqrt6\) m. This is the required result.
   4. Paving is charged per square metre (area) and edging per metre (perimeter). Paving: \(23\times18.90=£434.70\). Edging: \(20\times14.50=£290\). Total \(=434.70+290=\) £724.70. The total cost is £724.70.

8. Answer to Question 8

   1. Take out the factor \(-5\) from the terms in \(t\), then complete the square inside the bracket. \(h=-5(t^2-4t)+1.5=-5[(t-2)^2-4]+1.5=-5(t-2)^2+21.5\).
   2. Since \(-5(t-2)^2\le0\), \(h\) is largest when \(t-2=0\). Maximum height 21.5 m at \(t=\) 2 s. The ball reaches 21.5 m after 2 seconds.
   3. Use the quadratic formula on \(5t^2-20t-1.5=0\) and reject the negative time. \(5t^2-20t-1.5=0\). \(t=\dfrac{20\pm\sqrt{400+30}}{10}=\dfrac{20\pm20.736}{10}\). The positive root is \(t=\) 4.07 s (the other is negative). The ball hits the ground after about 4.07 s.
   4. Use the completed-square form to set up the inequality \(h>15\). \(-5(t-2)^2+21.5>15\) gives \((t-2)^2<1.3\), so \(2-\sqrt{1.3}\lt t<2+\sqrt{1.3}\). Duration \(=2\sqrt{1.3}=\) 2.28 s. The ball is higher than 15 m for about 2.28 s.

9. Answer to Question 9

   1. Profit is the profit per ticket multiplied by the number of tickets sold. Profit per ticket \(=x-3\), tickets sold \(=200-10x\), so \(P=(x-3)(200-10x)=200x-10x^2-600+30x=-10x^2+230x-600\). This is the required expression.
   2. Complete the square to find the maximum. \(P=-10(x^2-23x)-600=-10(x-11.5)^2+1322.5-600=-10(x-11.5)^2+722.5\). Greatest profit £722.50 at a price of £11.50 (85 tickets sold).
   3. Breaking even means \(P=0\); divide by \(-10\) and factorise. \(P=0\): \(x^2-23x+60=0\), \((x-3)(x-20)=0\), so £3 and £20. At these prices the school makes no profit.
   4. Use the completed-square form and solve the inequality. \(-10(x-11.5)^2+722.5\ge500\) gives \((x-11.5)^2\le22.25\), so \(|x-11.5|\le4.717\). Lower bound \(6.78\), upper \(16.22\). So prices from about £6.79 to £16.21 (rounded so the profit is still at least £500).

10. Answer to Question 10

    1. Use area \(=\tfrac12\times\text{base}\times\text{height}\). Area \(=\tfrac12\times4\sqrt3\times6=12\sqrt3\) m\(^2\). This is the required result.
    2. The height bisects the base of an isosceles triangle, giving a right-angled triangle; use Pythagoras. Half the base is \(2\sqrt3\). Side\(^2=(2\sqrt3)^2+6^2=12+36=48\), so the side is \(\sqrt{48}=\) \(4\sqrt3\) m. (So the sail is equilateral.)
    3. Find the area as a decimal, then multiply by the price. \(12\sqrt3=20.78\) m\(^2\). Cost \(=20.78\times12.40=257.7\approx\) £258. The cloth costs about £258.
    4. Areas scale with the square of lengths. \(\dfrac{48\sqrt3}{12\sqrt3}=4=2^2\), so lengths double. So each length must be multiplied by 2.

11. Answer to Question 11

    1. Two sides of length \(x\) use \(2x\) metres of fencing; the rest forms the side parallel to the wall. Length parallel to the wall \(=60-2x\), so \(A=x(60-2x)=60x-2x^2\). This is the required formula.
    2. Take out the factor \(-2\) and complete the square. \(A=-2(x^2-30x)=-2[(x-15)^2-225]=-2(x-15)^2+450\). Maximum area 450 m\(^2\) when \(x=15\) m and length \(=60-30=30\) m. So the pen should be 15 m wide and 30 m long.
    3. Use the completed-square form and solve the inequality. \(-2(x-15)^2+450\ge400\) gives \((x-15)^2\le25\), so \(10\le x\le20\). Any width from 10 m to 20 m gives at least 400 m\(^2\).
    4. \(-2(x-15)^2+450=500\) would need \((x-15)^2=-25\), which is impossible. The maximum area is only 450 m\(^2\). So 500 m\(^2\) cannot be enclosed.

12. Answer to Question 12

    1. Substitute each distance into \(T=r^{3/2}\). Mars: \(1.52^{1.5}=1.874\approx\) 1.87 years. Jupiter: \(5.2^{1.5}=11.858\approx\) 11.86 years. Mars takes about 1.87 years and Jupiter about 11.86 years to orbit the Sun.
    2. Raise both sides to the power \(\tfrac23\), the reciprocal of \(\tfrac32\). \(r^{3/2}=64\), so \(r=64^{2/3}=(\sqrt[3]{64})^2=4^2=\) 16 AU. The comet's mean distance is 16 AU.
    3. Use \(r=T^{2/3}\) with \(T=29.5\). \(r=29.5^{2/3}=9.55\) AU, very close to the tabulated 9.54. So the model fits Saturn very well.
    4. Square both sides and use \((a^m)^n=a^{mn}\). \(T^2=\left(r^{3/2}\right)^2=r^3\). This is Kepler's third law.

13. Answer to Question 13

    1. At an intersection the two \(y\)-values are equal. Equate: \(x^2-4x+7=2x+c\Rightarrow x^2-6x+7-c=0\). This is the required equation.
    2. The path just touches the bank when the quadratic has equal roots. Tangent means one repeated root, so the discriminant is zero: \(36-4(7-c)=0\), so \(8+4c=0\) and \(c=\) \(-2\). So \(c=-2\).
    3. Solve the equation with \(c=-2\). With \(c=-2\): \(x^2-6x+9=0\), \((x-3)^2=0\), \(x=3\) and \(y=2(3)-2=4\). Point \((3,4)\). The path touches the canal at \((3,4)\).
    4. Two crossing points need two distinct roots. Two crossing points need discriminant \(>0\): \(8+4c>0\), so \(c>-2\). So the path crosses the canal twice when \(c>-2\).

14. Answer to Question 14

    1. Since \(t=(\sqrt t)^2\), the expression is a quadratic in \(\sqrt t\). Let \(u=\sqrt t\): \(u^2-7u+10=0\), \((u-2)(u-5)=0\), so \(u=2\) or 5 and \(t=4\) or 25. Check: \(4-14+10=0\) ✓ and \(25-35+10=0\) ✓.
    2. The quadratic \(u^2-7u+10\) has a positive \(u^2\) coefficient, so it is negative between its roots. The expression is negative between its roots, so for \(4\lt t<25\) (weeks 5 to 24). The kiosk makes a loss between weeks 4 and 25.
    3. Complete the square in \(u\). In terms of \(u\): \(P=u^2-7u+10=(u-3.5)^2-2.25\). The minimum is \(-2.25\) when \(u=3.5\), i.e. \(t=12.25\). The greatest loss is \(2.25\times100=\) £225. The greatest weekly loss is £225, about 12 weeks after opening.

15. Answer to Question 15

    1. Substitute two values of \(x\) into \(A=x(75-x)\). \(x=30\): \(30\times45=1350\). \(x=60\): \(60\times15=900\). Both match the table.
    2. Expand and complete the square. \(A=-(x^2-75x)=-(x-37.5)^2+1406.25\). The maximum area is 1406.25 m\(^2\), achieved by a square of side 37.5 m.
    3. Set up the quadratic and use the formula. \(x(75-x)=1200\Rightarrow x^2-75x+1200=0\). Discriminant \(=5625-4800=825\). \(x=\dfrac{75\pm\sqrt{825}}2=\dfrac{75\pm28.72}2\), so \(x=\) 23.1 or 51.9. (The two fields are the same rectangle turned round.) The field could be 23.1 m by 51.9 m.
    4. Write the equation \(x(75-x)=1450\) and compute its discriminant. \(x^2-75x+1450=0\) has discriminant \(5625-5800=-175<0\), so no real solution: 1450 exceeds the maximum 1406.25. So an area of 1450 m\(^2\) is impossible.

16. Answer to Question 16

    1. The two shorter sides meet at the right angle, so use \(\tfrac12\times AB\times BC\) and the difference of two squares. Area \(=\tfrac12(\sqrt5+1)(\sqrt5-1)=\tfrac12(5-1)=2\) cm\(^2\). The surds cancel, so the area is exactly 2 cm\(^2\).
    2. Use Pythagoras: \(AC^2=AB^2+BC^2\), expanding each square carefully. \(AC^2=(\sqrt5+1)^2+(\sqrt5-1)^2=(6+2\sqrt5)+(6-2\sqrt5)=12\), so \(AC=\sqrt{12}=2\sqrt3\). So \(AC=2\sqrt3\) cm, as required.
    3. \(\tan A\) is the side opposite \(A\) (that is \(BC\)) divided by the side adjacent to \(A\) (that is \(AB\)); then rationalise the denominator. \(\tan A=\dfrac{BC}{AB}=\dfrac{\sqrt5-1}{\sqrt5+1}\). Multiply by \(\dfrac{\sqrt5-1}{\sqrt5-1}\): \(\dfrac{(\sqrt5-1)^2}{5-1}=\dfrac{6-2\sqrt5}4=\dfrac{3-\sqrt5}2\). This is the required result.
    4. Add the three side lengths; the \(+1\) and \(-1\) cancel. Perimeter \(=(\sqrt5+1)+(\sqrt5-1)+2\sqrt3=\) \(2\sqrt5+2\sqrt3\) cm. The perimeter is \((2\sqrt5+2\sqrt3)\) cm, about 7.94 cm.

17. Answer to Question 17

    1. Two distinct real roots need \(b^2-4ac>0\). \(36-4k>0\), so \(k<9\).
    2. For \(x^2+bx+c=0\), the sum of the roots is \(-b\) and the product is \(c\). Let the roots be \(\alpha\) and \(2\alpha\). Sum: \(3\alpha=6\), \(\alpha=2\). Product: \(2\alpha^2=8=k\). So \(k=8\) (roots 2 and 4). Check: \(2+4=6\) ✓ and \(2\times4=8\) ✓.
    3. \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=36-4k=16\), so \(k=5\). The equation \(x^2-6x+5=0\) has roots 1 and 5. Check: \(5-1=4\) ✓.
    4. Sum of roots \(=6>0\) and product \(=k>0\). A positive product means the roots have the same sign, and a positive sum means that sign is positive. Real roots need \(k\le9\). So for \(0\lt k<9\) both roots are positive.

18. Answer to Question 18

    1. 8 and 4 are both powers of 2, so write both sides with base 2 and compare the indices. Write both sides as powers of 2: \(2^{3(x-1)}=2^{2(x+2)}\), so \(3x-3=2x+4\) and \(x=\) 7. Check: \(8^6=2^{18}=4^9\) ✓.
    2. Write every term in terms of \(3^x\). Let \(y=3^x\). Then \(9^x=y^2\) and \(3^{x+1}=3y\). \(y^2-12y+27=0\), \((y-3)(y-9)=0\), so \(3^x=3\) or \(9\) and \(x=\) 1 or 2.
    3. Raise both sides to the power \(\tfrac23\). \(x=125^{2/3}=(\sqrt[3]{125})^2=5^2=\) 25. Check: \(25^{3/2}=5^3=125\) ✓.
    4. Write \(4^{x-1}\) as \(2^{2x-2}\) and use the division law. \(\dfrac{2^{x+3}}{2^{2x-2}}=2^{x+3-2x+2}=2^{5-x}\). So \(2^{5-x}=2^4\) and \(x=\) 1. Check: \(\dfrac{2^4}{4^0}=16\) ✓.

19. Answer to Question 19

    1. Substitute each value of \(x\) from the table into the formula. \(x=5\): \(-0.08(25)+8=6\). \(x=10\): \(-8+16=8\). \(x=15\): \(-0.08(225)+24=6\). \(x=0\) and \(x=20\): 0 (\(-0.08\times400+32=0\)). All match. So the model fits all the data.
    2. The boat is centred at \(x=10\), so it occupies 3 m on each side; the lowest point of the arch above the boat is at its edges. The boat occupies \(7\le x\le13\). The arch is lowest at the edges: \(h(7)=-0.08\times49+11.2=7.28>6.5\). So the boat fits with 0.78 m to spare.
    3. Write \(h=-0.08(x-10)^2+8\) and set \(h=7\). \(-0.08(x-10)^2+8=7\) gives \((x-10)^2=12.5\), so \(x=10\pm3.536\). Width \(=2\times3.536=\) 7.07 m. At a height of 7 m the arch is about 7.07 m wide.
    4. The widest boat at height 7 m has width 7.07 m, which is the answer to part (c) (the same calculation). So a 7 m high boat can be at most 7.07 m wide.

20. Answer to Question 20

    1. The roots are real and distinct exactly when the discriminant is positive. Discriminant \(=k^2-4(k-2)=k^2-4k+8=(k-2)^2+4\). Since \((k-2)^2\ge0\), the discriminant is at least 4, so it is always positive. Two distinct real roots exist for every \(k\).
    2. For \(x^2+bx+c=0\) the sum of the roots is \(-b\) and the product is \(c\). \(\alpha+\beta=-k\) and \(\alpha\beta=k-2\). \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=k^2-2(k-2)=k^2-2k+4=(k-1)^2+3\). The minimum value is 3, when \(k=1\).
    3. Both roots negative needs sum \(<0\) and product \(>0\). Sum: \(-k<0\) so \(k>0\). Product: \(k-2>0\) so \(k>2\). Both conditions hold when \(k>2\). So both roots are negative when \(k>2\).