Solving Algebra and Linear and Quadratic Equations (Coursework Sample)

Student’s Name
Professor’s Name
Course Number
Date
MATH TASK 1 - SOLUTIONS
Problem 1
x2-xy+8=0………………………(1)
x2-8x+y=0………………………(2)
Subtract eqn. 2 from eqn. 1
8x-xy+8-y=0
x8-y+1(8-y)=0
x+1+(8-y)=0
x=-1, y=8
Substitute x=-1 in eqn. 1
(-1)2-(-1)y+8=0
1+y+8=0
y=-9
Substitute y=8 in eqn. 2
x2-8x+8=0
x=-b∓b2-4ac2a
x=8∓64-4*1*82*1
x=8∓322
x=8∓2*162
x=8∓4*22
x=4∓22
x=22-2, x=22+2
x=22-2, y=8
x=22+2, y=8
x=-1, y=8
Problem 2
At the intersect with the x-axis, y=0, therefore,
(x-1)2-4=0
(x-1)2-22=0
Factorizing the LHS of the above equation we have:
x-1-2*x-1+2=0
x-1-2*x-1+2=0
(x-3)(x+1)=0
x=-1, x=3
The x-axis intersects then are -1,0and (3,0)
The mid-pointa,b=-1+32,0
a,b=1,0
Therefore, a=1
Problem 3
Part a
Let the point through which all parabolas of the form y=x2+2ax+a passes be (x1, y1) and let m and n be the two distinct real numbers, then,
y1=x12+2mx1+m…………………………(1)
y1=x12+2nx1+n………………………...(2)
If (x1, y1) is common to the above two parabolas, then RHS of equation 1 is equal to RHS of equation 2:
x12+2mx1+m=x12+2nx1+n
2mx1-2nx1=-m+n
2x1(m-n)=-(m-n)
x1=-(m-n)2(m-n)
x1=-12
Replace x1=-12 in equation 2
y1=(-12)2+2*-12n+n
y1=14-n+n
y1=14
Therefore the coordinates through which all the parabolas pass is -12,14
Part b
Finding the vertex of each parabola y=x2+2ax+a,
At the vertex of the above parabola, the gradient is equal to zero meaning, dydx=0
dydy=2x+2a=0
x=-a
Replacingx=-a, to get the value ofy,
y=-a2+2a-a+a
y=a2-2a2+a
y=a-a2
Therefore the vertex has coordinates, -a, a-a2
Thus the vertices lies on a parabola with equation
y=-x2-x
For this parabola then, at its vertex, dydx=0,
dydx=-2x-1=0
x=-12
Then,
y=--122--12
y=14
Therefore the vertex of this curve is -12,14 which the same common point is as the parabolas of the formy=x2+2ax+a.
Problem 4
Part a: Sketch the graph of the equation y=xx-42
First find the y – intercept where x=0
y=00-42
y=0
Therefore the y – intercept is (0,0)
Find the x- intercept wherey=0,
xx-42=0
x=0, x-42=0
x=0, x=4
Therefore the y – intercept is 0,0and 4,0
Find the turning points (minimum and maximum points) of the equation.
At the turning points, dydx=0
dydx=xddxx-42+x-42ddxx=0
dydx=2x-4x+x-42*1=0
3x2-16x+16=0
Solving for x
x=4, x=43
Replacing the values of x to get y we have the following turning points:
Maxuimum 43,25627, Minimum 4, 0
Getting a point beyond the minimum at x = 5, y = 5
Getting a point beyond 0 to the left where x=-1, y=-25
We therefore have the following points to plot:
x

y

-1

25

0

0

43

25627

4

0

5

5

Below is the plot of y=xx-42
Part b: Solve the inequality:
xx-42≥0
x≥0, x-42≥0
x-42≥0
x≥4
Therefore x≥0 is the solution of the inequality.
Problem 5
p+pr+pr2=26…………………………………(1)
p2r+p2r2+p2r3=156………………………….(2)
Simplifying Equation 2
prp+pr+pr2=156…………………………..(3)
Replace p+pr+pr2 in equation 3 with 26 from equation 1
26pr=256
pr=6
p=6r
Replace p=26r in equation 1
6r+6r*r+6r*r2=26
6r+6+6r=26
6+6r+6r2=26r
6r2-20r+6=0
6r2-18r-2r+6=0
6rr-3-2r-3=0
6r-2r-3=0
r=12,r=3
Forr=12, p=18
For r=3 p=2
p=18, r=12p=2, r=3
Problem 6:
Given a quadratic equation ax2+bx+c=0, if a, b and c are consecutive terms in a geometric sequence then we can take the values as,
a=pr, b=p, c=pr
Where p≠0 and p, r>0
Now replacing the values of a, b and c in the quadratic equation we will have:
prx2+px+pr=0
x2+rx+r2=0
From the above equation, the discriminant is,
∆=b2-4ac
∆=r2-4*1*r2
∆=-3r2
From the above we see that the discriminant has no real roots and hence the coefficients a, b and c of the quadratic equation cannot be consecutive terms in a geometric progression.
Problem 7
Given a quadratic equation ax2+bx+c=0, if a, b and c are consecutive terms in an arithmetic progression then let the integer roots of the quadratic equation be α,β
ax2+bx+c=a[x2+bax+ca]
ax2+bax+ca=ax-αx-β as α,β are the two roots
ax2+bax+ca=ax2-α+βx+αβ
Then
ba=-(α+β)ca=αβ