Algebra Questions with Solutions and Answers for Grade 11

Algebra Questions with Solutions and Answers for Grade 11

algebra questions with answers and detailed solutions, for grade 11, are presented.

    Questions

  1. Complete the square in the quadratic function f given by
    f(x) = 2x2 - 6x + 4

  2. Find the point(s) of intersection of the parabola with equation y = x2 - 5x + 4 and the line with equation y = 2x - 2

  3. Find the constant k so that : -x2 - (k + 7)x - 8 = -(x - 2)(x - 4)

  4. Find the center and radius of the circle with equation x2 + y2 -2x + 4y - 11 = 0

  5. Find the constant k so that the quadratic equation 2x2 + 5x - k = 0 has two real solutions.

  6. Find the constant k so that the system of the two equations: 2x + ky = 2 and 5x - 3y = 7 has no solutions.

  7. Factor the expression 6x2 - 13x + 5

  8. Simplify i231 where i is the imaginary unit and is defined as: i = √(-1).

  9. What is the remainder when f(x) = (x - 2)54 is divided by x - 1?

  10. Find b and c so that the parabola with equation y = 4x2 - bx - c has a vertex at (2 , 4)?

  11. Find all zeros of the polynomial P(x) = x3 - 3x2 - 10x + 24 knowing that x = 2 is a zero of the polynomial.

  12. If x is an integer, what is the greatest value of x which satisfies 5 < 2x + 2 < 9?

  13. Sets A and B are given by: A = {2 , 3 , 6 , 8, 10} , B = {3 , 5 , 7 , 9}.
    a) Find the intersection of sets A and B.
    b) Find the union of sets A and B.

  14. Simplify | - x2 + 4x - 4 |.

  15. Find the constant k so that the line with equation y = kx is tangent to the circle with equation (x - 3)2 + (y - 5)2 = 4.

Solutions to the Above Questions


  1. f(x) = 2(x2 - 3x) + 4 : factor 2 out in the first two terms
    = 2(x2 - 3x + (-3/2)2 - (-3/2)2) + 4 : add and subtract (-3/2)2
    = 2(x - 3/2))2 - 1/2 : complete square and group like terms

  2. 2x - 2 = x2 - 5x + 4 : substitute y by 2x - 2
    x = 1 and x = 6 : solution of quadratic equation
    (1 , 0) and (6 , 10) : points of intersection

  3. -x2 - (k + 7)x - 8 = -(x - 2)(x - 4) : given
    -x2 - (k + 7)x - 8 = -x2 + 6x - 8
    -(k + 7) = 6 : two polynomials are equal if their corresponding coefficients are equal.
    k = -13 : solve the above for k

  4. x2 - 2x + y2 + 4y = 11 : Put terms in x together and terms in y together
    (x - 1)2 + (y + 2)2 - 1 - 4 = 11
    (x - 1)2 + (y + 2)2 = 42 : write equation of circle in standard form
    center(1 , -2) and radius = 4 : identify center and radius

  5. 2x2 + 5x - k = 0 : given
    discriminant = 25 - 4(2)(-k) = 25 + 8k
    25 + 8k > 0 : quadratic equations has 2 real solutions when discriminant is positive
    k > -25/8

  6. Determinant = -6 - 5k
    -6 - 5k = 0 : when determinant is equal to zero (and equations independent) the system has no solution
    k = -6/5 : solve for k

  7. 6x2 - 13x + 5 = (3x - 5)(2x - 1)

  8. Note that i4 = 1
    Note also that 231 = 4 * 57 + 3
    Hence i231 = (i4)57 * i3
    = 157 * -i = -i

  9. remainder = f(1) = (1 - 2)54 = 1 : remainder theorem

  10. h = b / 8 = 2 : formula for x coordinate of vertex
    b = 16 : solve for b
    y = 4 for x = 2 : the vertex point is a solution to the equation of the parabola
    4(2)2 - 16(2) - c = 4
    c = -20 : solve for c

  11. divide P(x) by (x - 2) to obtain x2 - x + 12
    P(x) = (x2 - x + 12)(x - 2)
    = (x - 4)(x + 3)(x - 2) : factor the quadratic term
    the zeros are : 4 , -3 and 2

  12. 5 < 2x + 2 < 9 : given
    3/2 < x < 7/2
    the greatest integer value of is 3 (the integer less than 7/2)

  13. A intersection B = {3} : common element to both A and B is 3
    A union B = {2 , 3 , 6 , 8, 10 , 5 , 7 , 9} : all elements of A and B are in the union. Elements common to both A and B are listed once only since it is a set.

  14. | - x2 + 4x - 4 | : given
    = | -(x2 + 4x - 4) |
    = | -(x - 2)2 |
    = (x - 2)2

  15. (x - 3)2 + (y - 5)2 = 4 : given
    (x - 3)2 + (kx - 5)2 = 4 : substitute y by kx
    x2(1 + k2) - x(6 + 10k) + 21 = 0 : expand and write quadratic equation in standard form.
    (6 + 10k)2 - 4(1 + k2)(21) = 0 : For the circle and the line y = kx to be tangent, the discriminant of the above quadratic equation must be equal to zero.
    16k2 + 120k - 48 = 0 : expand above equation
    k = (-15 + √(273)/4 , k = (-15 - √(273)/4 : solve the above quadratic equation.

More References and links

High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers
Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers
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