Algebra Questions with Answers and Solutions - Grade 12

Solutions to the Above Problems

1. 
   25¹⁰⁰
   2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰
   3⁴⁰⁰ = (3⁴)¹⁰⁰ = 81¹⁰⁰
   4²⁰⁰ = (4²)¹⁰⁰ = 16¹⁰⁰
   2⁶⁰⁰ = (2⁶)¹⁰⁰ = 64¹⁰⁰ from greatest to least: 3⁴⁰⁰ , 2⁶⁰⁰, 25¹⁰⁰ , 4²⁰⁰ , 2³⁰⁰
   
2. 
   P(x) = x³ - 7x + 6 : given
   leading coefficient 1 and its factors are : +1,-1
   constant term is 6 and its factors are : +1,-1,+2,-2,+3,-3,+6,-6
   possible rational zeros : +1,-1,+2,-2,+3,-3,+6,-6
   test : P(1) = 0, P(2) = 0 and P(-3) = 0
   x = 1, x = 2 and x = -3 are the zeros of P(x).
   
3. 
   From the graph, x = -3 is a zero of multiplicity 2, x = 0 is a zero of multiplicity 1 and x = 2 is a zero of multiplicity 2.
   P(x) = -x (x + 3)²(x - 2)² : polynomial with real zeros hence with lowest degree.
   
4. )
   if 2 - i is a zero and the coefficients of the polynomial are real then 2 + i (the conjugate) is also a solution.
   P(x) = (x - (2 - i))(x - (2 + i))×q(x) = ((x - 2)² + 1)×q(x)
   q(x) = P(x)/((x - 2)² + 1) = (x² - 2)
   x = 2 - i , x = 2 + i , x = √2 and x = - √2 are the 4 zeros of P(x).
   
5. 
   f(x) = a (x + 2)² + 1 : equation of parabola in vertex form
   f(0) = -3 = 4 a + 1
   a = -1 : solve for a
   f(x) = -(x + 2)² + 1 = -x² - 4x - 3
   a = -1 , b = -4 and c = -3 : identify coefficients
   
6. 
   f(x) = a x² + b x + c
   f(1) = 3 which give 3 = a + b + c
   f(5) = 3 which gives 3 = 25a + 5b + c
   24 a + 4 b = 0 : subtract equation B from equation C
   x coordinate of vertex = -b/2a = 3 : from above equation
   
7. 
   The oblique asymptote is the quotient resulting from the long division of ax⁴ + bx³ + 3 by x³ - 2
   The quotient obtained is ax + b
   a x + b = 2 x - 3
   a = 2 and b = -3 : for two polynomials to be equal, the corresponding coefficients has to be equal.
   
8. 
   log₉ (x³) = log₂(8) : given
   log₂ (2³) = 3 : simplify right hand side of given equation.
   log₉ (x³) = 3 : rewrite the above equation
   log₉ (x³) = log₉(9³) : rewite 3 as a log base 9.
   x³ = 9³ : obtain algebraic equation from eqaution D.
   x = 9 : solve above for x.
   
9. 
   log_x (y³) = 2 : given
   x² = y³ : rewrite in exponential form
   x⁴ = y⁶ : square both sides
   x⁴ = y⁶ : rewrite the above using the log base y
   log_y(x⁴) = log_y(y⁶) = 6
   
10. 
    log_x (8 e³) = 3 : given
    x³ = 8 e³ = (2e)³
    x = 2 e
    
11. 
    16^x + 16^{x - 1} = 10 : given
    4^2x + 4^2x / 16 = 10
    4^2x = 160/17 : solve for 4^2x
    4^x = 4 √(10) / √(17) : extract the square root
    2^2x = 4^x = 4 √(10) / √(17)
    
12. 
    a² - b² = 8 : given
    a⁴ + b⁴ - 2a²b² = 8² : square both sides and expand.
    a×b = 2 : given
    a²b² = 2² : square both sides.
    a⁴ + b⁴ - 2(4) = 8² : substitute
    a⁴ + b⁴ = 72
    
13. 
    -1 ? sin(2x - ?/3) ? 1 : range of a sine function
    -2 ? 2sin(2x - ?/3) ? 2 : multiply all terms of the double inequality by 2
    -2 - 5 ? 2sin(2x - ?/3) - 5 ? 2 - 5 : add -5 to all terms of the inequality.
    -7 ? 2sin(2x - ?/3) - 5 ? -3
    3 ? |2sin(2x - ?/3) - 5| ? 7 : change the above using absolute value.
    3 + 3 ? |2sin(2x - ?/3) - 5| + 3 ? 7 + 3 : add 3 to all terms of the double inequality.
    The maximum value of f(x) is equal to 10 and the minimum value of f(x) is equal to 6.
    
14. 
    If x < -7 then x < - 3 and x + 3 < 0 and |3 + x| = -(3 + x)
    |4 - |3 + x|| = |4 + 3 + x| = |x + 7| = - (x + 7) = - x - 7 : since x + 7 < 0
    
15. 
    Let d be the distance between A and B
    T1 = d / 50 : travel time from A to B
    Let S be the speed from B to A
    T2 = d/S : travel time from B to A
    60 = 2d/(T1 + T2) : average speed for the whole trip
    60 = 2d/(d/50 + d/S) : substitute T1 and T2
    S = 75 km/hour : solve the above equation for S.
    
16. 
    x² - y² = (x - y)(x + y) = -12 : given
    6(x - y) = -12 : substitute x + y by 6
    (x - y) = -2 : solve for x - y
    (x - y) = -2 and x + y = 6 : 2 by 2 system.
    x = 2 , y = 4 : solve above system.
    
17. 
    f(x) + 3f(8 - x) = x : given
    f(2) + 3f(6) = 2 : x = 2 above
    f(6) + 3f(2) = 6 : x = 6 above
    f(6) = 6 - 3f(2) : solve equation C for f(6)
    f(2) + 3(6 - 3f(2)) = 2 : substitute
    f(2) = 2 : solve above equation.
    
18. 
    f(2x + 1) = 2f(x) + 1 : given
    f(3) = 2f(1) + 1 : x = 1 in A
    f(1) = 2f(0) + 1 : x = 0 in A
    f(3) = 11 : substitute
    
19. 
    x² + y² = 4 : given
    x² + (2x + b)² = 4 : substitute y by 2x + b
    5x² + 4bx + b² - 4 = 0
    The number of points of intersection is given by the number of solutions of the above equation. The line and circle are tangent if the above quadratic equation has only one solution which means that the discriminant is equal to zero. Find the discriminant as a function of b and solve.
    b = √2 and b = -√2 : 2 solutions.
    
20. 
    (x¹⁰⁰ - x⁹⁹ - x + 1) / (x² - 3x + 2)
    Let P(x) = x¹⁰⁰ - x⁹⁹ - x + 1 , D(x) = x² - 3x + 2
    The division of the two polynomials may be written as
    P(x) = D(x) Q(x) + r(x) , where Q(x) is the quotient and r(x) is the remainder that will have a degree equal to one or lower. r(x) = a x + b
    We now need to find a and b that define the remainder.
    Note that D(x) may be factored as follows: D(x) = x² - 3x + 2 = (x - 1)(x - 2)
    Hence: P(x) = (x - 1)(x - 2) Q(x) + a x + b
    Using the zeros of D(x) to write:
    P(1) = (1 - 1)(1 - 2) Q(1) + a (1) + b gives a + b = P(1)
    P(2) = (2 - 1)(2 - 2) Q(2) + a (2) + b gives 2 a + b = P(2)
    We now need to evaluate P(1) and P(2)
    P(1) = 1¹⁰⁰ - 1⁹⁹ - (1) + 1 = 0
    First rewrite P(x) = x⁹⁹(x - 1) - x + 1 ; Hence P(2) = 2⁹⁹(2 - 1) - 2 + 1 = 2⁹⁹ - 1
    We now have a system of equations to solve and find a and b.
    a + b = 0 and 2 a + b = 2⁹⁹ - 1
    a = 2⁹⁹ - 1 and b = 1 - 2⁹⁹
    remainder: r(x) = (2⁹⁹ - 1) x + 1 - 2⁹⁹
    
21. 
    Let y = √(1/3 + √(1/3 + √(1/3 + ...))).
    square both sides to obtain: y ² = 1/3 + √(1/3 + √(1/3 + √(1/3 + ...)))
    We can write: y ² = 1/3 + y
    Solve the above quadratic equation to obtain: y = (3 + √(21)) / 6 and y = (3 - √21) / 6
    y is positive hence the solution: √(1/3 + √(1/3 + √(1/3 + ...))) = y = (3 + √(21)) / 6

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