# Solving Algebra and Linear and Quadratic Equations (Coursework Sample)

included a number of math problems in algebra, linear equations and quadratic equations

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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

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