Algebra Questions with Solutions and Answers for Grade 11
algebra questions with answers and detailed solutions, for grade 11, are presented.
Questions
Complete the square in the quadratic function f given by
f(x) = 2x2 - 6x + 4
Find the point(s) of intersection of the parabola with equation y = x2 - 5x + 4 and the line with equation y = 2x - 2
Find the constant k so that : -x2 - (k + 7)x - 8 = -(x - 2)(x - 4)
Find the center and radius of the circle with equation x2 + y2 -2x + 4y - 11 = 0
Find the constant k so that the quadratic equation 2x2 + 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 6x2 - 13x + 5
Simplify i231 where i is the imaginary unit and is defined as: i = √(-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 = 4x2 - bx - c has a vertex at (2 , 4)?
Find all zeros of the polynomial P(x) = x3 - 3x2 - 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 | - x2 + 4x - 4 |.
Find the constant k so that the line with equation y = k x is tangent to the circle with equation (x - 3)2 + (y - 5)2 = 4.
Solutions to the Above Questions
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
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
-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
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
2x2 + 5x - k = 0 : given
discriminant = 25 - 4(2)(-k) = 25 + 8k
25 + 8k > 0 : quadratic equations has 2 real solutions when the discriminant is positive
k > -25/8
Determinant = -6 - 5k
-6 - 5k = 0 : when the determinant is equal to zero (and equations independent) the system has no solution
k = -6/5 : solve for k
6x2 - 13x + 5 = (3x - 5)(2x - 1)
Note that i4 = 1
Note also that 231 = 4 * 57 + 3
Hence i231 = (i4)57 * i3 = 157 * -i = -i
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 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
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 that are common to both A and B are listed once only since it is a set.
(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 the 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.
