Grade 11Ac Mathematics Assignment 7
Due Date
A. In the diagram below, 3 straight line graphs and
1 parabola are drawn. Lines A and B are parallel.
𝒚
C.
Each row, column and 3x3 diagonal in the square
below adds up to 3.
D
𝑥−4
𝑥+1
𝑥
B
A
−𝟐
𝒙
𝟑
𝟐
𝟏
−𝟑
𝑥+𝑦
C
𝟓
3𝑥 − 2𝑦
Which pair of simultaneous equations has:
Equation pair
a solution with a positive 𝒙-value
and negative 𝒚-value ?
A
C
no solution ?
A
B
a solution with 𝒚-value 𝟎 ?
B
C
two distinct solutions ?
C
D
one repeated root solution ?
B
D
B. Use the Completing the Square method to solve the
following quadratic equations, leaving your answers
in surd form:
(a)
𝑥 2 + 6𝑥 − 23 = 0
(b)
𝑥 2 + 8𝑥 + 4 = 0
(c)
𝑥 2 − 3𝑥 − 2 = 0
2𝑥
𝟎
−𝟏
𝟒
(a) Find the value of 𝑥 and 𝑦
From top row deduce that 𝒙 = 𝟐.
Using 𝑥 = 2 and left side column, deduce that 𝒚 = 𝟑.
(b) Complete the table with possible expressions
using 𝑥 and 𝑦 for each empty square.
Any expressions that correctly equate to the missing
values, shown in blue in the square above.
D. A paint mixture is made up of blue (b) and yellow
(y) and paint and must satisfy:
2𝑏 + 𝑦 < 14 , 𝑏 ≥ 3,
𝑦>4
(a) Sketch the inequalities on the grid below.
𝒚
9
(See next page)
𝟏𝟒
A ball is shot vertically up into the air from ground
level. The ball’s distance (in meters) above ground
level, 𝒔, is given by the formula:
𝒔 = 𝟐𝟒. 𝟓𝒕 − 𝟒. 𝟗𝒕𝟐
where 𝑡 is time, in seconds.
(d) At what times will the ball be exactly 19.6 m
above ground level?
19.6 = 24.5𝑡 − 4.9𝑡 2
4.9𝑡 2 − 24.5𝑡 + 19.6 = 0
÷ 4.9 to give … 𝑡 2 − 5𝑡 + 4 = 0
(𝑡 − 4)(𝑡 − 1) = 0
So, 𝒕 = 𝟒 or 𝒕 = 𝟏 seconds
Self Reflection:
I got
𝟒
𝟑
𝒃
(b) If the paint comes in bottles of 1 liter only, list all
the possible combinations of blue and yellow 1L paint
bottles that may be used to make the paint mixture.
3 blue and 5 yellow, 3 blue and 6 yellow
3 blue and 7 yellow, 4 blue and 5 yellow
I did best at
correct.
𝟕
My target for improvement is
Question B
(a)
(b)
𝑥 2 + 6𝑥 − 23 = 0
𝑥 2 + 8𝑥 + 4 = 0
𝑥 2 + 6𝑥 = 23
8 2
8 2
𝑥 2 + 8𝑥 + ( ) = −4 + ( )
2
2
6 2
6 2
𝑥 2 + 6𝑥 + ( ) = 23 + ( )
2
2
𝑥 2 + 8𝑥 + 16 = −4 + 16
𝑥 2 + 6𝑥 + 9 = 23 + 9
(𝑥 + 4)2 = 12
(𝑥 + 3)2 = 32
𝑥 = −4 ± 2√3
𝒙 = −𝟑 ± 𝟒√𝟐
𝒙 = 𝟐(−𝟐 ± √𝟑)
(c)
𝑥 2 − 3𝑥 −
9
=0
2
3 2 9
3 2
𝑥 2 − 3𝑥 + ( ) = + ( )
2
2
2
3 2 27
(𝑥 − ) =
2
4
𝑥=
3
27 3 3√3
±√ = ±
2
4
2
2
𝒙=
𝟑
(𝟏 ± √𝟑)
𝟐