Indices, Logarithms, Polynomials and Quadratic Equations

Algebra manipulation questions are the backbone of the BCA paper. Three of the 50 questions come from here, but the same skills are needed in sequences, matrices and calculus, so a weak spot costs you in other chapters too.

The Dean's Office model asks the nature of roots of x^2-8x+16=0, a root-to-constant question, and simple equation word problems ("product of two consecutive even numbers is 288").

4.1 Laws of indices and surds

a^ma^n=a^{m+n}, \frac{a^m}{a^n}=a^{m-n}, (a^m)^n=a^{mn}, (ab)^n=a^nb^n, a^0=1, a^{-n}=\frac{1}{a^n}, a^{m/n}=\sqrt[n]{a^m}. If a^x=a^y and a\ne0,\pm1 then x=y. Factor common powers: 2^{x+2}-2^x=2^x(4-1)=3\cdot2^x.

A surd such as \sqrt{5} is an irrational root; (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b rationalises \frac{1}{\sqrt{a}+\sqrt{b}}.

4.2 Logarithms and their laws

\log_a x=y\iff a^y=x (a>0, a\ne1, x>0). Laws: \log(mn)=\log m+\log n; \log\frac{m}{n}=\log m-\log n; \log m^p=p\log m; \log_a a=1; \log_a1=0; \log_ab=\frac{1}{\log_ba}=\frac{\log b}{\log a}; a^{\log_ax}=x.

Standard values: \log_{10}2\approx0.3010, \log_{10}3\approx0.4771, \log_{10}5=1-\log_{10}2\approx0.6990. \log(m+n) has no expansion.

(a) John Napier, inventor of logarithms

(b) A slide rule: logarithmic scales

(c) Slide rule: add lengths to multiply

4.3 Polynomials, remainder and factor theorem

• Remainder theorem: f(x)\div(x-a) leaves remainder f(a). Factor theorem: (x-a) is a factor iff f(a)=0.
• (a\pm b)^2=a^2\pm2ab+b^2; a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2); if a+b+c=0 then a^3+b^3+c^3=3abc.
• If x+\frac{1}{x}=k then x^2+\frac{1}{x^2}=k^2-2.

4.4 Quadratic equations, nature and sum and product of roots

For ax^2+bx+c=0: roots \frac{-b\pm\sqrt{D}}{2a} with D=b^2-4ac.

D	Roots
D>0, perfect square	real, unequal, rational
D>0, not a square	real, unequal, irrational
D=0	real and equal
D<0	imaginary (complex conjugates)

\alpha+\beta=-\frac{b}{a}, \alpha\beta=\frac{c}{a}, \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta. The equation with roots \alpha,\beta is x^2-(\alpha+\beta)x+\alpha\beta=0.

Roots are opposite if b=0 and reciprocal if a=c. ax^2+bx+c has extreme value -\frac{D}{4a} at x=-\frac{b}{2a}.

4.5 Linear and simultaneous equations, inequalities, equation word problems

• Eliminate one variable; for x+y=s, x-y=d: x=\frac{s+d}{2}, y=\frac{s-d}{2}.
• Consecutive integers: n,n+1,\dots; consecutive even: n,n+2,\dots.
• Inequalities: multiplying or dividing by a negative reverses the sign. |x|<a\iff-a<x<a; |x|>a\iff x<-a or x>a.
• Radical equations: square both sides, then check every root (extraneous roots appear).

4.6 Eight traps in indices, logarithms, polynomials and equations

Algebra marks are lost to small habits, not to hard ideas. The eight traps below are the ones that appear again and again in the Dean's Office model, and each can be checked with one line of arithmetic.

1. a^0=1, and a negative index means a reciprocal, not a negative number. So 2^{-2}=\frac14, not -4 and not -\frac12 (Fig 4.1). Also (-2)^2=4 but -2^2=-4, because the power acts before the minus sign. And (a+b)^n is never a^n+b^n: (1+2)^2=9 but 1^2+2^2=5.
2. Fractional powers are roots, and surds do not add inside. a^{m/n} means take the n-th root and then raise to m, so 8^{2/3}=2^2=4 and 27^{-1/3}=\frac13. The sum \sqrt{a}+\sqrt{b} is not \sqrt{a+b}: \sqrt9+\sqrt{16}=7 but \sqrt{25}=5. To remove a surd from the denominator multiply by the conjugate, as in \frac{1}{\sqrt3-1}=\frac{\sqrt3+1}{2}.
3. The logarithm of a sum is not the sum of logarithms. The law is \log(ab)=\log a+\log b for a product only, so \log(2+3)\ne\log2+\log3 (Fig 4.2). Remember \log_a1=0, \log_aa=1, and the argument must be positive. A question that says "log" without a base means base 10, while \ln means base e.
4. Know the three working laws and the change of base. \log\frac ab=\log a-\log b, \log a^n=n\log a, and \log_ab=\frac{\log b}{\log a}, so \log_ab\times\log_ba=1. Also a^{\log_ax}=x. Typical key results are \log_264=6 and \log5=1-\log2=0.699, because 5=\frac{10}{2}.
5. Use the root of the divisor in the remainder theorem. The remainder of p(x) divided by (x-a) is p(a), so for (x+2) you substitute x=-2, and for (2x-1) you substitute x=\frac12 (Fig 4.3). If p(a)=0 then (x-a) is a factor. Do not divide by x to "simplify" before you check whether x=0 is itself a root.
6. Read a, b, c with their signs, then use -\frac ba and \frac ca. For x^2-5x+6=0 the sum of roots is -\frac{-5}{1}=5 and the product is 6, giving roots 2 and 3. The discriminant D=b^2-4ac decides the nature (Fig 4.4): D>0 two real roots, D=0 equal roots, D<0 none. A quadratic with given roots \alpha,\beta is x^2-(\alpha+\beta)x+\alpha\beta=0.
7. Never cancel a factor that can be zero, and check radical roots. From x^2=3x the answers are x=0 and x=3; dividing by x throws away one. Squaring an equation with a square root can create a false root, so substitute back. In inequalities, multiplying or dividing by a negative number reverses the sign, and |x|<a means -a<x<a.
8. Two equations give a point, no point, or a whole line. Compare \frac{a_1}{a_2}, \frac{b_1}{b_2} and \frac{c_1}{c_2}: all different slopes gives one solution, equal ratios of a and b but not c gives none, and all three equal gives infinitely many (Fig 4.5). In word problems define each letter in a sentence first, solve, then test the answer against the story, such as ages that must be positive whole numbers.

4.6.1 Worked checks

Q. Given \log2=0.3010 and \log3=0.4771, find \log 12 and \log 1.5.

Answer. \log12=\log(2^2\times3)=2(0.3010)+0.4771=1.0791. \log1.5=\log\frac32=\log3-\log2=0.4771-0.3010=0.1761. Both come from the product, power and quotient laws, never from adding inside the log.

Q. Find the remainder when p(x)=x^3-6x^2+11x-6 is divided by (x-4), and say whether (x-2) is a factor.

Answer. Remainder =p(4)=64-96+44-6=6, which synthetic division in Fig 4.3 also gave. For (x-2): p(2)=8-24+22-6=0, so yes, (x-2) is a factor. Checking the other roots, p(1)=0 and p(3)=0 too.

Q. For 2x^2-7x+3=0 find the nature of the roots, the roots, and verify the sum and product.

Answer. D=49-24=25>0, so the roots are real and unequal. x=\frac{7\pm5}{4} gives x=3 and x=\frac12. Sum =\frac72=-\frac{b}{a} and product =\frac32=\frac ca, which agree.

Chapter summary

• Reduce to a common base before equating indices.
• \log(m+n) cannot be split; \log(mn) can.
• D decides the nature of roots; D=0 gives equal roots.
• Use \alpha+\beta and \alpha\beta instead of solving for the roots.
• f(a) is the remainder on division by (x-a).
• A negative multiplier reverses an inequality.
• Always substitute back after squaring.

Practice questions

Set A — Recall (20 questions)

1. The value of 2^3\times2^4 isa64b32c128d256
2. The value of \log_{10}1000 isa3b10c100d1
3. For x\ne0, x^0 equalsa0bxcundefinedd1
4. The sum of the roots of x^2-5x+6=0 isa6b5c-5d1
5. The product of the roots of 2x^2-7x+3=0 isa-\frac{3}{2}b\frac{7}{2}c\frac{3}{2}d\frac{2}{3}
6. The discriminant of x^2-6x+9=0 isa36b18c-36d0
7. \log_a(mn) equalsa\log_am+\log_anb\log_am\cdot\log_anc\log_am-\log_andn\log_am
8. The value of 8^{2/3} isa2b4c\frac{16}{3}d8
9. For the equation x^2-8x+16=0 the roots areaimaginary and unequalbrational and unequalcirrationaldreal and equal
10. The remainder when x^2+3x+2 is divided by (x-1) isa0b2c6d1
11. \log_aa equalsa1b0cadundefined
12. The value of (\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3}) isa8b2c15d\sqrt{2}
13. If x-5=1, then x+7 equalsa5b7c13d12
14. (x+y)^2 equalsax^2+y^2bx^2+xy+y^2cx^2-2xy+y^2dx^2+2xy+y^2
15. The roots of x^2-9=0 area3, -3b9, -9c3 onlyd0, 9
16. If 2^x=32, then x isa4b5c6d16
17. If one root of x^2-7x+c=0 is 3, then c isa10b21c12d4
18. The value of \log_381 isa4b27c3d9
19. The solution of -2x>6 isax>-3bx>3cx<3dx<-3
20. |x|<2 meansax<2b-2<x<2cx<-2 or x>2dx>-2

Answers and explanations — Set A

1. (c) — 2⁷ = 128; multiplying powers adds exponents (not 2¹² = 4096 either).

2. (a) — 10³ = 1000.

3. (d) — any non-zero number to the power 0 is 1.

4. (b) — −b/a = 5.

5. (c) — c/a = 3/2.

6. (d) — 36 − 36 = 0, a perfect square.

7. (a) — product becomes a sum.

8. (b) — 8^{2/3} = (∛8)² = 4.

9. (d) — D = 64 − 64 = 0.

10. (c) — f(1) = 1 + 3 + 2 = 6.

11. (a) — a¹ = a.

12. (b) — a² − b² pattern: 5 − 3 = 2.

13. (c) — x = 6, so x + 7 = 13.

14. (d) — the middle term is +2xy.

15. (a) — x² = 9 has two roots, ±3.

16. (b) — 32 = 2⁵.

17. (c) — 9 − 21 + c = 0, so c = 12.

18. (a) — 3⁴ = 81.

19. (d) — dividing by −2 reverses the sign: x < −3.

20. (b) — |x| < 2 is a band around zero; (c) is |x| > 2.

Set B — Understanding (12 questions)

1. If 3^{x+1}=81, then x isa2b4c1d3
2. Given \log_{10}2=0.301, the value of \log_{10}8 isa0.602b0.903c1.204d0.301
3. If \alpha,\beta are the roots of x^2-5x+6=0, then \alpha^2+\beta^2 isa13b25c37d11
4. If the roots of x^2+px+q=0 are 2 and 3, then p+q isa11b-1c1d5
5. The quadratic equation whose roots are 2 and -3 isax^2-x-6=0bx^2+5x+6=0cx^2-5x-6=0dx^2+x-6=0
6. kx^2-6x+9=0 has equal roots when k equalsa1b2c4d9
7. (x-2) is a factor of x^3-3x^2+kx+2. Then k isa0b1c2d-1
8. If x+\frac{1}{x}=3, then x^2+\frac{1}{x^2} isa5b6c7d9
9. The product of two consecutive even numbers is 288. Their sum isa34b30c32d36
10. The sum of two numbers is 15 and their difference is 3. The larger isa8b9c10d12
11. If 2^{x+2}-2^x=12, then x isa1b3c4d2
12. The solution of \sqrt{x+6}=x isa-2b6c3d3 and -2

Answers and explanations — Set B

1. (d) — 81 = 3⁴, so x + 1 = 4.

2. (b) — log 8 = 3 log 2 = 0.903.

3. (a) — 5² − 2·6 = 13.

4. (c) — p = −5, q = 6, so p + q = 1.

5. (d) — sum −1, product −6, so x² + x − 6.

6. (a) — D = 36 − 36k = 0.

7. (b) — f(2) = 8 − 12 + 2k + 2 = 0 gives k = 1.

8. (c) — 3² − 2 = 7.

9. (a) — 16 × 18 = 288; sum 34.

10. (b) — (15 + 3)/2 = 9.

11. (d) — 2ˣ(4 − 1) = 12, so 2ˣ = 4.

12. (c) — x² − x − 6 = 0 gives 3 or −2, but −2 fails since √4 ≠ −2.

Set C — Application (8 questions)

1. The value of 1^3+2^3+(-3)^3 isa-18b-6c0d18
2. The minimum value of x^2-6x+11 isa11b3c0d2
3. The sum of the squares of two consecutive positive integers is 85. Their product isa30b35c42d56
4. If x-y=2 and x+y=10, then xy isa20b24c16d32
5. If x+\frac{1}{x}=\frac{5}{2}, the values of x area5 and \frac{1}{5}b3 and \frac{1}{3}c1 and 4d2 and \frac{1}{2}
6. The value of \log_28+\log_39 isa5b6c17d4
7. The roots of x^2-5x+k=0 differ by 1. Then k isa4b5c6d8
8. How many integers x satisfy x>-1 and 3x+2<14?a3b4c5d2

Answers and explanations — Set C

1. (a) — a + b + c = 0, so the sum of cubes is 3abc = 3(1)(2)(−3) = −18.

2. (d) — (x − 3)² + 2, minimum 2.

3. (c) — 6² + 7² = 85, product 42.

4. (b) — x = 6, y = 4, xy = 24.

5. (d) — 2x² − 5x + 2 = 0 gives x = 2 or 1/2.

6. (a) — 3 + 2 = 5.

7. (c) — roots 2 and 3, so k = 6.

8. (b) — x < 4 and x > −1 gives 0, 1, 2, 3.