Algebra Questions with Solutions and Answers for Grade 11
Grade 11 math algebra questions with answers and solutions are presented.
a

Complete the square in the quadratic function f given by
f(x) = 2x^{2}  6x + 4

Find the point(s) of intersection of the parabola with equation y = x^{2}  5x + 4 and the line with equation y = 2x  2

Find the constant k so that : x^{2}  (k + 7)x  8 = (x  2)(x  4)

Find the center and radius of the circle with equation x^{2} + y^{2} 2x + 4y  11 = 0

Find the constant k so that the quadratic equation 2x^{2} + 5x  k = 0 has two real solutions.

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

Factor the expression 6x^{2}  13x + 5

Simplify i^{231} where i is the imaginary unit and is defined as: i = sqrt(1).

What is the remainder when f(x) = (x  2)^{54} is divided by x  1?

Find b and c so that the parabola with equation y = 4x^{2}  bx  c has a vertex at (2 , 4)?

Find all zeros of the polynomial P(x) = x^{3}  3x^{2}  10x + 24 knowing that x = 2 is a zero of the polynomial.

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

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.

Simplify   x^{2} + 4x  4 .

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 Problems


f(x) = 2(x^{2}  3x) + 4 : factor 2 out in the first two terms
= 2(x^{2}  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


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


x^{2}  (k + 7)x  8 = (x  2)(x  4) : given

x^{2}  (k + 7)x  8 = x^{2} + 6x  8
(k + 7) = 6 : two polynomials are equal if their corresponding coefficients are equal.
 k = 13 : solve the above for k


x^{2}  2x + y^{2} + 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} = 4^{2} : write equation of circle in standard form
 center(1 , 2) and radius = 4 : identify center and radius


2x^{2} + 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

 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


6x^{2}  13x + 5 = (3x  5)(2x  1)

 Note that i<4> = 1
 Note also that 231 = 4 * 57 + 3

Hence i^{231} = (i^{4})^{57} * i^{3}

= 1^{57} * i = i


remainder = f(1) = (1  2)^{54} = 1 : remainder theorem

 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


divide P(x) by (x  2) to obtain x^{2}  x + 12

P(x) = (x^{2}  x + 12)(x  2)
 = (x  4)(x + 3)(x  2) : factor the quadratic term
 the zeros are : 4 , 3 and 2

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

 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.


  x^{2} + 4x  4  : given

=  (x^{2} + 4x  4) 

=  (x  2)^{2} 

= (x  2)^{2}


(x  3)^{2} + (y  5)^{2} = 4 : given

(x  3)^{2} + (kx  5)^{2} = 4 : substitute y by kx

x^{2}(1 + k^{2})  x(6 + 10k) + 21 = 0 : expand and write quadratic equation in standard form.

(6 + 10k)^{2}  4(1 + k^{2})(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.

16k^{2} + 120k  48 = 0 : expand above equation
 k = (15 + sqrt(273)/4 , k = (15  sqrt(273)/4 : solve the above quadratic equation.


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