Algebra and Graphs

Question

Solve x squared + 5x - 14 = 0.

Solution

1. 1

   Find two numbers that multiply to -14 and add to +5

   Factors of -14: (7, -2) since 7 x (-2) = -14 and 7 + (-2) = 5

   For x squared + bx + c = 0, we need two numbers that multiply to c and add to b.

2. 2

   Write the factorised form

   (x + 7)(x - 2) = 0

   Place each number into a bracket with x.

3. 3

   Solve each bracket

   x + 7 = 0, so x = -7. Or x - 2 = 0, so x = 2.

   If two things multiply to give zero, at least one must be zero.

Question

Solve: 3x + 2y = 16 and 5x - 2y = 24.

Solution

1. 1

   Add the equations to eliminate y

   3x + 2y + 5x - 2y = 16 + 24, so 8x = 40

   Since one equation has +2y and the other -2y, adding eliminates y.

2. 2

   Solve for x

   x = 40 / 8 = 5

   Divide both sides by 8.

3. 3

   Substitute to find y

   3(5) + 2y = 16, so 15 + 2y = 16, 2y = 1, y = 0.5

   Substitute x = 5 back into either original equation.

4. 4

   Check in the other equation

   5(5) - 2(0.5) = 25 - 1 = 24. Correct.

   Always verify your answer in the equation you did not use.

Question

Simplify (x squared - 9) / (x squared + 5x + 6).

Solution

1. 1

   Factorise the numerator

   x squared - 9 = (x + 3)(x - 3)

   This is a difference of two squares: a squared - b squared = (a+b)(a-b).

2. 2

   Factorise the denominator

   x squared + 5x + 6 = (x + 2)(x + 3)

   Find two numbers that multiply to 6 and add to 5: that is 2 and 3.

3. 3

   Cancel common factors

   (x + 3)(x - 3) / (x + 2)(x + 3) = (x - 3) / (x + 2)

   (x + 3) appears in both numerator and denominator so it cancels.

Question

Solve 2x squared - 3x - 7 = 0. Give answers to 2 decimal places.

Solution

1. 1

   Identify a, b, c

   a = 2, b = -3, c = -7

   Compare with ax squared + bx + c = 0.

2. 2

   Substitute into the quadratic formula

   x = (-b plus or minus sqrt(b squared - 4ac)) / 2a = (3 plus or minus sqrt(9 + 56)) / 4

   x = (-(-3) plus or minus sqrt((-3) squared - 4(2)(-7))) / (2 times 2)

3. 3

   Calculate the discriminant

   b squared - 4ac = 9 + 56 = 65

   The discriminant tells us there are two real solutions since 65 > 0.

4. 4

   Find both solutions

   x = (3 + sqrt(65)) / 4 = (3 + 8.062) / 4 = 11.062 / 4 = 2.77. Or x = (3 - 8.062) / 4 = -5.062 / 4 = -1.27.

   Calculate each solution separately using plus then minus.

Question

f(x) = 3x + 1 and g(x) = x squared. Find fg(2).

Solution

1. 1

   Apply g first (inner function)

   g(2) = 2 squared = 4

   fg(2) means f(g(2)). Apply g first.

2. 2

   Apply f to the result

   f(4) = 3(4) + 1 = 13

   Now substitute g(2) = 4 into f.

Question

f(x) = (2x - 5) / 3. Find f inverse of x.

Solution

1. 1

   Write y = f(x)

   y = (2x - 5) / 3

   Replace f(x) with y.

2. 2

   Swap x and y

   x = (2y - 5) / 3

   This is the key step for finding inverses.

3. 3

   Solve for y

   3x = 2y - 5, then 3x + 5 = 2y, then y = (3x + 5) / 2

   Multiply both sides by 3, add 5, divide by 2.

Question

f(x) = 2x + 3 and g(x) = x squared - 1. Find (a) fg(x), (b) gf(x).

Solution

1. 1

   Find fg(x)

   fg(x) = f(g(x)) = f(x squared - 1) = 2(x squared - 1) + 3 = 2x squared - 2 + 3 = 2x squared + 1

   Replace every x in f with the expression g(x) = x squared - 1.

2. 2

   Find gf(x)

   gf(x) = g(f(x)) = g(2x + 3) = (2x + 3) squared - 1 = 4x squared + 12x + 9 - 1 = 4x squared + 12x + 8

   Replace every x in g with the expression f(x) = 2x + 3.

Question

f(x) = x squared - 4x + 3 for 0 <= x <= 5. Find the range of f.

Solution

1. 1

   Find the minimum point

   x = -b/2a = 4/2 = 2. f(2) = 4 - 8 + 3 = -1

   The minimum of a quadratic ax squared + bx + c occurs at x = -b/(2a).

2. 2

   Find the values at the endpoints

   f(0) = 0 - 0 + 3 = 3. f(5) = 25 - 20 + 3 = 8.

   Check the function value at both ends of the domain.

3. 3

   State the range

   The minimum value is -1 (at x = 2) and the maximum is 8 (at x = 5). Range: -1 <= f(x) <= 8.

   The range includes all output values from the minimum to the maximum.

Question

Find the nth term of the sequence 7, 11, 15, 19, 23, ...

Solution

1. 1

   Find the common difference

   11-7 = 4, 15-11 = 4. Common difference d = 4.

   Subtract consecutive terms to find the common difference.

2. 2

   Write the nth term

   nth term = 4n + (7-4) = 4n + 3

   nth term = dn + (first term - d).

3. 3

   Verify

   n=1: 4(1)+3 = 7. n=2: 4(2)+3 = 11. n=3: 4(3)+3 = 15. All correct.

   Always check your formula against the given terms.

Question

Find the nth term of the sequence 3, 9, 19, 33, 51, ...

Solution

1. 1

   Find first and second differences

   First differences: 6, 10, 14, 18. Second differences: 4, 4, 4.

   Constant second difference means it is a quadratic sequence.

2. 2

   Find the coefficient of n squared

   Coefficient of n squared = second difference / 2 = 4/2 = 2. So the sequence starts with 2n squared.

   For a quadratic sequence an squared + bn + c, a = second difference / 2.

3. 3

   Subtract 2n squared from each term

   n=1: 3-2=1, n=2: 9-8=1, n=3: 19-18=1, n=4: 33-32=1, n=5: 51-50=1

   The remainders form a constant sequence.

4. 4

   Write the nth term

   nth term = 2n squared + 1

   Since the remainders are all 1, the linear part is 0n + 1.

Question

Find the sum of the first 20 terms of the sequence 5, 8, 11, 14, ...

Solution

1. 1

   Identify a and d

   First term a = 5, common difference d = 3

   These are needed for the sum formula.

2. 2

   Use the sum formula

   S(n) = n/2 (2a + (n-1)d) = 20/2 (2(5) + 19(3)) = 10(10 + 57) = 10(67) = 670

   Substitute n = 20, a = 5, d = 3 into the formula.

Question

Solve 3x - 7 > 2x + 5.

Solution

1. 1

   Collect x terms on one side

   3x - 2x > 5 + 7

   Subtract 2x from both sides and add 7 to both sides.

2. 2

   Simplify

   x > 12

   Combine like terms.

Question

Solve -3 < 2x + 1 <= 9 and list the integer values.

Solution

1. 1

   Subtract 1 from all parts

   -3 - 1 < 2x <= 9 - 1, so -4 < 2x <= 8

   Apply the same operation to all three parts.

2. 2

   Divide all parts by 2

   -2 < x <= 4

   Divide each part by 2.

3. 3

   List integer values

   x can be -1, 0, 1, 2, 3, 4

   -2 is not included (strict inequality) but 4 is included (<=).

Question

On a grid, show the region satisfying y >= 1, x + y <= 6, and y <= 2x.

Solution

1. 1

   Draw y = 1

   Horizontal line through y = 1. Shade ABOVE (y >= 1). Solid line.

   y >= 1 means all points on or above the line y = 1.

2. 2

   Draw x + y = 6

   Line through (0,6) and (6,0). Shade BELOW/LEFT (x+y <= 6). Solid line.

   Test (0,0): 0+0 = 0 <= 6, true, so shade the side containing (0,0).

3. 3

   Draw y = 2x

   Line through (0,0) and (1,2). Shade BELOW/RIGHT (y <= 2x). Solid line.

   Test (3,1): 1 <= 6, true, so shade the side containing (3,1).

4. 4

   Identify the region

   The required region is the triangle where all three conditions are satisfied simultaneously.

   The feasible region is the overlap of all three shaded regions.

Question

Solve x squared - 5x + 6 <= 0.

Solution

1. 1

   Factorise

   x squared - 5x + 6 = (x-2)(x-3) = 0 when x = 2 or x = 3

   First find where the expression equals zero.

2. 2

   Determine the solution region

   The parabola y = x squared - 5x + 6 opens upwards (positive coefficient of x squared). It is negative (below the x-axis) between the roots.

   For <= 0 with an upward parabola, the solution is between the roots.

3. 3

   Write the solution

   2 <= x <= 3

   Include the endpoints because the inequality is <=.

Question

Find dy/dx for y = 3x cubed - 5x squared + 2x - 7.

Solution

1. 1

   Differentiate each term using the power rule

   d/dx(3x cubed) = 9x squared. d/dx(-5x squared) = -10x. d/dx(2x) = 2. d/dx(-7) = 0.

   Multiply the coefficient by the power, then reduce the power by 1. Constants become 0.

2. 2

   Combine

   dy/dx = 9x squared - 10x + 2

   Write all the differentiated terms together.

Question

Find the stationary points of y = x cubed - 6x squared + 9x + 1 and determine their nature.

Solution

1. 1

   Differentiate

   dy/dx = 3x squared - 12x + 9

   Use the power rule on each term.

2. 2

   Set dy/dx = 0

   3x squared - 12x + 9 = 0. Divide by 3: x squared - 4x + 3 = 0. Factorise: (x-1)(x-3) = 0. x = 1 or x = 3.

   Stationary points occur where the gradient is zero.

3. 3

   Find y-coordinates

   When x = 1: y = 1 - 6 + 9 + 1 = 5. When x = 3: y = 27 - 54 + 27 + 1 = 1.

   Substitute x-values back into the original equation.

4. 4

   Find second derivative and classify

   d2y/dx2 = 6x - 12. At x = 1: d2y/dx2 = 6 - 12 = -6 < 0, so (1, 5) is a maximum. At x = 3: d2y/dx2 = 18 - 12 = 6 > 0, so (3, 1) is a minimum.

   Positive second derivative = minimum, negative = maximum.

Question

Find the equation of the tangent to y = x squared - 3x + 5 at the point where x = 2.

Solution

1. 1

   Find the y-coordinate

   y = 4 - 6 + 5 = 3. So the point is (2, 3).

   Substitute x = 2 into the original equation.

2. 2

   Find the gradient at x = 2

   dy/dx = 2x - 3. At x = 2: gradient = 2(2) - 3 = 1.

   Differentiate and substitute x = 2 to find the gradient of the tangent.

3. 3

   Write the tangent equation

   y - 3 = 1(x - 2), so y = x + 1.

   Use y - y1 = m(x - x1) with the point and gradient.

Question

Differentiate y = 4/x squared + 3 sqrt(x). Write in index form first.

Solution

1. 1

   Rewrite using index notation

   y = 4x to the power -2 + 3x to the power 1/2

   1/x squared = x to the power -2, and sqrt(x) = x to the power 1/2.

2. 2

   Differentiate each term

   dy/dx = 4(-2)x to the power -3 + 3(1/2)x to the power -1/2 = -8x to the power -3 + (3/2)x to the power -1/2

   Apply the power rule to each term.

3. 3

   Rewrite if needed

   dy/dx = -8/x cubed + 3/(2 sqrt(x))

   Convert back to fraction form if the question requires it.