Algebra — SSC CGL Quant Notes

Concepts and rules

(a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b). Also (a + b)² − (a − b)² = 4ab and (a + b)² + (a − b)² = 2(a² + b²).

(a + b)³ = a³ + b³ + 3ab(a + b) and (a − b)³ = a³ − b³ − 3ab(a − b).

a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²).

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca), and a² + b² + c² − ab − bc − ca = 1/2 × [(a − b)² + (b − c)² + (c − a)²].

Consequences: if a + b + c = 0, then a³ + b³ + c³ = 3abc. If a³ + b³ + c³ = 3abc for real numbers, then either a + b + c = 0 or a = b = c.

Reciprocal forms: if x + 1/x = k, then x² + 1/x² = k² − 2 and x³ + 1/x³ = k³ − 3k. If x − 1/x = k, then x² + 1/x² = k² + 2 and x³ − 1/x³ = k³ + 3k.

For real x > 0, x + 1/x ≥ 2, with equality only at x = 1; for x < 0, x + 1/x ≤ −2. A sum of squares of real numbers is 0 only if each of them is 0.

Method and shortcuts

Step 1: Look at what is given (sum, product, x + 1/x) and pick the identity that links it to the expression asked for.

Step 2: Build up powers in order: from a + b and ab get a² + b², then a³ + b³; from x + 1/x get x² + 1/x², then x⁴ + 1/x⁴ = (x² + 1/x²)² − 2.

Step 3: In numerical expressions spot the pattern: (a³ + b³)/(a² − ab + b²) = a + b and (a³ − b³)/(a² + ab + b²) = a − b.

Shortcut: when a + b + c = 0, replace a³ + b³ + c³ by 3abc at once. For symmetric expressions, put simple values that satisfy the condition (for example a = 1, b = 1, c = −2) and test the options.

Value testing fails when different values satisfying the condition give different results; test two different sets of values before trusting it.

Squares and products near a base: (100 + a)² = 10000 + 200a + a² and (100 + a)(100 − a) = 10000 − a².

Example 1: Sum and product given

Question: If a + b = 10 and ab = 21, find a² + b² and a³ + b³.

Solution: a² + b² = (a + b)² − 2ab = 100 − 42 = 58. a³ + b³ = (a + b)³ − 3ab(a + b) = 1000 − 3 × 21 × 10 = 1000 − 630 = 370.

Check: a and b are 3 and 7, and 9 + 49 = 58, 27 + 343 = 370.

Answer: a² + b² = 58 and a³ + b³ = 370.

Example 2: x + 1/x given

Question: If x + 1/x = 5, find x² + 1/x² and x³ + 1/x³.

Solution: x² + 1/x² = 5² − 2 = 25 − 2 = 23. x³ + 1/x³ = 5³ − 3 × 5 = 125 − 15 = 110.

Answer: 23 and 110.

Example 3: Fast squares and products

Question: Find 997² and 103 × 97 without long multiplication.

Solution: 997² = (1000 − 3)² = 1000000 − 6000 + 9 = 994009. 103 × 97 = (100 + 3)(100 − 3) = 10000 − 9 = 9991.

Answer: 994009 and 9991.

Example 4: Three variables

Question: If a + b + c = 12 and a² + b² + c² = 50, find a³ + b³ + c³ − 3abc. (a) 72 (b) 36 (c) 48 (d) 24

Solution: 144 = 50 + 2(ab + bc + ca), so ab + bc + ca = 47.

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) = 12 × (50 − 47) = 36.

Check with a = 3, b = 4, c = 5, which fit both conditions: 27 + 64 + 125 − 3 × 60 = 216 − 180 = 36.

Answer: (b) 36

Example 5: Surd value of x

Question: If x = 2 + √3, find x³ + 1/x³. (a) 56 (b) 64 (c) 48 (d) 52

Solution: 1/x = 1/(2 + √3) = (2 − √3)/[(2 + √3)(2 − √3)] = (2 − √3)/(4 − 3) = 2 − √3.

So x + 1/x = 4, and x³ + 1/x³ = 4³ − 3 × 4 = 64 − 12 = 52. The option 64 is the trap of forgetting the − 3k term.

Answer: (d) 52

Example 6: Going down from the fourth power

Question: If x⁴ + 1/x⁴ = 119 and x > 1, find x³ − 1/x³. (a) 36 (b) 27 (c) 33 (d) 39

Solution: (x² + 1/x²)² = x⁴ + 1/x⁴ + 2 = 121, so x² + 1/x² = 11, taking the positive root because it is a sum of squares.

(x − 1/x)² = x² + 1/x² − 2 = 9, so x − 1/x = 3; it is positive because x > 1 makes x greater than 1/x.

x³ − 1/x³ = (x − 1/x)³ + 3(x − 1/x) = 27 + 9 = 36.

Answer: (a) 36

Common mistakes

Writing (a + b)² = a² + b². The middle term 2ab is missing; it is zero only if a or b is 0.

Using x² + 1/x² = k² + 2 when x + 1/x = k. It is k² − 2 for the sum and k² + 2 for the difference.

Taking a square root without checking the sign: from (x − 1/x)² = 9, x − 1/x may be 3 or −3. Use the given condition (such as x > 1) to choose.

Using a³ + b³ + c³ = 3abc without first checking that a + b + c = 0.

Rationalising carelessly: 1/(2 + √3) = 2 − √3 only because (2 + √3)(2 − √3) = 1; for 1/(3 + √5) the denominator becomes 9 − 5 = 4.

Practice questions

Q1. If a − b = 4 and ab = 45, find a³ − b³. (a) 540 (b) 604 (c) 476 (d) 644

Q2. If a + b + c = 0, a = 3 and b = −5, find a³ + b³ + c³. (a) 90 (b) −30 (c) −90 (d) 30

Q3. Simplify (0.73³ + 0.27³)/(0.73² − 0.73 × 0.27 + 0.27²). (a) 0.46 (b) 0.27 (c) 0.73 (d) 1

Answers

Q1. (b) a³ − b³ = (a − b)³ + 3ab(a − b) = 64 + 3 × 45 × 4 = 64 + 540 = 604. Check: a = 9, b = 5 gives 729 − 125 = 604.

Q2. (c) c = 2 and the sum is 0, so a³ + b³ + c³ = 3abc = 3 × 3 × (−5) × 2 = −90. Check: 27 − 125 + 8 = −90.

Q3. (d) The form is (a³ + b³)/(a² − ab + b²) = a + b = 0.73 + 0.27 = 1.

Quick revision

a² + b² = (a + b)² − 2ab; a³ + b³ = (a + b)³ − 3ab(a + b); a³ − b³ = (a − b)³ + 3ab(a − b).

x + 1/x = k: squares k² − 2, cubes k³ − 3k. x − 1/x = k: squares k² + 2, cubes k³ + 3k.

a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc.

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

After any square root, fix the sign from the condition given.

Concepts and rules

An equation ax + by + c = 0, with a and b not both 0, is a linear equation in two variables. Its graph is a straight line, and every point on the line is a solution.

The slope of the line through (x1, y1) and (x2, y2) is (y2 − y1)/(x2 − x1). For ax + by + c = 0 with b ≠ 0, slope = −a/b and y-intercept = −c/b; with a ≠ 0, x-intercept = −c/a.

y = mx + c has slope m and y-intercept c. Two lines are parallel when their slopes are equal, and perpendicular when the product of their slopes is −1 (neither line being vertical).

x = k is a vertical line, parallel to the y-axis; y = k is a horizontal line, parallel to the x-axis. The x-axis itself is y = 0 and the y-axis is x = 0.

For a1x + b1y = c1 and a2x + b2y = c2 (a2, b2, c2 non-zero): if a1/a2 ≠ b1/b2 there is exactly one solution (the lines cross); if a1/a2 = b1/b2 ≠ c1/c2 there is no solution (parallel lines); if a1/a2 = b1/b2 = c1/c2 there are infinitely many (the same line).

The line ax + by = c with a, b, c all non-zero meets the axes at (c/a, 0) and (0, c/b), and the triangle it makes with the axes has area c²/(2|ab|).

Area of the triangle with vertices (x1, y1), (x2, y2), (x3, y3) = 1/2 × |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|.

Method and shortcuts

Step 1: Solve a pair of equations by elimination: make the coefficients of one variable equal, then add or subtract the equations.

Step 2: For a triangle formed by lines, first find every corner (where the lines meet each other and the axes), then use base × height/2 or the vertex formula.

Step 3: For consistency questions, write the three ratios a1/a2, b1/b2 and c1/c2, get k from the two ratios that must be equal, then check the third ratio.

Cross-multiplication: for a1x + b1y = c1 and a2x + b2y = c2, x = (c1b2 − c2b1)/(a1b2 − a2b1) and y = (a1c2 − a2c1)/(a1b2 − a2b1), provided a1b2 − a2b1 ≠ 0.

Shortcut: to test an option, substitute the point in both equations; any option that fails one equation is out.

Example 1: Solving a pair

Question: Solve 3x + 2y = 12 and x − y = −1.

Solution: Multiply the second equation by 2: 2x − 2y = −2. Adding it to the first gives 5x = 10, so x = 2, and then y = x + 1 = 3.

Check: 3 × 2 + 2 × 3 = 12 and 2 − 3 = −1.

Answer: x = 2, y = 3.

Example 2: Triangle with the axes

Question: Find the area of the triangle formed by the line 4x + 3y = 24 and the coordinate axes.

Solution: Put y = 0 to get (6, 0) and x = 0 to get (0, 8). Area = 1/2 × 6 × 8 = 24. Formula check: 24²/(2 × 4 × 3) = 576/24 = 24.

Answer: 24 square units.

Example 3: No solution

Question: For what value of k do kx + 6y = 9 and 2x + 3y = 5 have no solution?

Solution: Need k/2 = 6/3 ≠ 9/5. From k/2 = 2, k = 4; and 9/5 is not 2, so the lines are parallel, not the same line.

Answer: k = 4.

Example 4: Infinitely many solutions

Question: For what value of k do 2x + 3y = 7 and (k − 1)x + (k + 2)y = 3k have infinitely many solutions? (a) 5 (b) 6 (c) 7 (d) 8

Solution: Need 2/(k − 1) = 3/(k + 2) = 7/(3k). From the first two, 2(k + 2) = 3(k − 1), so 2k + 4 = 3k − 3 and k = 7.

Check the third ratio: 3/(k + 2) = 3/9 = 1/3 and 7/(3k) = 7/21 = 1/3. All three ratios are equal, so the two equations describe the same line.

Answer: (c) 7

Example 5: Triangle with two lines and the y-axis

Question: Find the area of the triangle formed by the lines 2x + y = 8, x − y = 1 and the y-axis. (a) 12 (b) 13.5 (c) 27 (d) 9

Solution: Adding the equations gives 3x = 9, so the lines meet at (3, 2). On the y-axis (x = 0) the first line gives y = 8 and the second gives y = −1.

The base on the y-axis runs from −1 to 8, length 9; the height is the distance of (3, 2) from the y-axis, which is 3.

Area = 1/2 × 9 × 3 = 13.5 square units.

Answer: (b) 13.5

Example 6: Corners from three conditions

Question: The line 3x − 4y = 12 meets the x-axis at A and the y-axis at B. The line through B parallel to the x-axis meets the line x + y = 10 at C. Find the area of triangle ABC. (a) 19.5 (b) 39 (c) 18 (d) 21

Solution: y = 0 gives A(4, 0); x = 0 gives y = −3, so B(0, −3).

The line through B parallel to the x-axis is y = −3; with x + y = 10 it gives x = 13, so C(13, −3).

BC lies on y = −3 and has length 13; the height from A to this line is 0 − (−3) = 3. Area = 1/2 × 13 × 3 = 19.5 square units.

Answer: (a) 19.5

Common mistakes

Getting the sign of the slope wrong: for ax + by + c = 0 the slope is −a/b, not a/b.

Thinking that a1/a2 = b1/b2 alone means infinitely many solutions. The third ratio decides between no solution and infinitely many.

Not checking the third ratio after finding k. If it also matches in a no-solution question, that k gives the same line, not parallel lines.

Using a negative intercept as a negative length. Lengths and areas use absolute values.

Calling x = 3 parallel to the x-axis. x = 3 is vertical and parallel to the y-axis.

Practice questions

Q1. What is the slope of the line through (2, 3) and (6, 11)? (a) 1/2 (b) 2 (c) −2 (d) 4

Q2. For what value of k do 3x + ky = 6 and 6x + 8y = 12 have infinitely many solutions? (a) 2 (b) 6 (c) 8 (d) 4

Q3. The sum of a two-digit number and the number formed by reversing its digits is 121, and the tens digit exceeds the units digit by 3. What is the number? (a) 47 (b) 83 (c) 74 (d) 65

Answers

Q1. (b) Slope = (11 − 3)/(6 − 2) = 8/4 = 2.

Q2. (d) 3/6 = k/8 = 6/12 = 1/2, so k = 4.

Q3. (c) Let the tens digit be a and the units digit b. (10a + b) + (10b + a) = 121 gives a + b = 11; with a − b = 3, a = 7 and b = 4, so the number is 74.

Quick revision

Slope = (y2 − y1)/(x2 − x1) = −a/b; parallel lines have equal slopes, perpendicular lines have product −1.

Ratios: a1/a2 ≠ b1/b2 one solution; a1/a2 = b1/b2 ≠ c1/c2 none; all equal infinitely many.

Triangle of ax + by = c with the axes: c²/(2|ab|).

Find every corner before computing an area; x = k is vertical and y = k is horizontal.