Quadratic Functions and the Substitution Method (Math Problem Sample)

Mathematic Assignment 0004
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Question 1 Answer
Let the sides of the triangle be a,b,c
a=b-1 and=b+1 and b=b
r=△s=ss-as-bs-cs=4
s=3b2
b2+1b2b2-13b2=4
Square both side
b2+1b2b2-13b2=16
Simplify the equation
b+22b2b-223b2
(b2-4)b=192b
Which implies that b3-4b=192b
Which implies that b3=196b
Divide both side by b ,
b2=196 which implies that b=14
Then a=13 ,b=14 ,c=15
Question 2 Answer
Rewrite the equation x2+x-3=0 as quadratic function y=x2+x-3 . Draw the graph of y=x2+x-3 for -3≤x≤2
x

-3

-2

-1

0

1

2

y

3

-1

-3

-3

-1

3

The solution for the equation y=x2+x-3 can be obtained by looking at the points where the graph y=x2+x-3 cuts the x-axis.
The graphy=x2+x-3, cuts they -axis at x  1.3 and x  –2.3
So, the solution for the equation x2+x-3 is x  1.3 or x  –2.3.
Question 3 Answer
Solve the simultaneous equation by substitution method.
Get the values of y. Then, the values of x.
3x2-5xy+2y2=33………. i
4x-3y=14……………. ii
Make x to be the subject of the formular using equation (ii)
x=14+3y4
Substitute x to equation (i) above.
3(14+3y4)-5(14+3y4)y+2y2=33
Multiply both sides by 4
42+9y-70-15y+8y2=165
After substituting, the equation becomes
8y2-6y-193
Find the value of y using quadratic formula.
y=6±62+4×8×1932×8
It follows that y is eq