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Quadratic Functions.............................

Find all fixed points for the following functions by setting the function equal to x and solving the resulting equation for x
 
(1) f(x)=x^2-9x+25
 
Application of quadratic equation and quadratic functions
 
(1) Two consecutive positive odd integers have a product of 99,find the integers.
 
(2) The Product of two consecutive odd positive integers is 14 more than three times the larger integer.Find the the integers.
 
 
 
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2 Answers

1) x^2 - 9x + 25 = x
 
subract x from both sides
x^2 - 9x + 25 = x-x
X^2 - 10x +25 = 0
 
its a quadratic equation of general form Ax^2 + Bx + C = 0
 
If you learn the following procedure you will be able to solve any quadratic equation:
 
Trick is to:
 
(a) split the coefficient of x (B) in two numbers such that sum of these two numbers is B , AND
(b) product of these two split numbers must be equal to product of coefficient of X^2 and constant  i.e A x C
(c) rewrite qaudratic equation and factorize
 
If this doesn not work the you can apply the following following formula for solution of a quadratic equation:
 
x = (- B +/- sqrt ( B^2 - 4AC))/2A
 
In this question A =1, B=-10, C=25
 
(a) split B into two  i.e -5, -5 such that -5 -5 = -10
(b) check product of these two = A x C; -5 x -5 = 1 x 25
Now rewrite quadratic equation
 
x^2 - 5x - 5x +25 = 0
x(x - 5) -5(x - 5)
(x-5)(x-5) = 0
(x-5)^2 = 0
x-5=0
x=5
 
Applying formula will always get you the answer with factorizing but may take more time
 
x = (-(-10) +/- sqrt ( (-10)^2 - 4x1x25C))/2x1
x= (10 +/- sqrt (100-100))/2
x= (10 +/-(0))/2
x= (10+0)/2 or (10-0)/2= 5 :)
 
(2) n(n+2) =99
 
n^2 + 2n -99 = 0
 
Split 2 into 11 and -9 so that sum is +2 and product is -99
rewrite
 
n^2 + 11n - 9n -99 =0
n(n+11) - 9(n+11) = 0
(n+11)(n-9) = 0
n = -11 or +9 since we are looking for positive integers answer is 11 and 9
 
Apply formula :
 
Compare with general quadratic equation Ax^2 + Bx + C = 0
 
A= 1, B= 2 and C= -99
 
x = (-2 +/- sqrt ( 2^2 - (4x1x(-99)))/2x1
X= (-2 +/- sqrt (4- (-369)))/2
x= (-2 +/- sqrt(400))/2
x= (-2 +/- 20)2
x= (-2+20)/2 or (-2-20)/2
x= 9 or -11 :)
 
Good luck with the second application
1) x^2-9x+25 = x
=> x^2-10x+25 = (x-5)^2 = 0
Answer: x = 5
 
Application
1) n(n+2) = 99
=> n^2+2n-99 = (n+11)(n-9) = 0
n = 9
Answer: 9, 11
 
2) n(n+2) = 3(n+2)+14
=> n^2-n-20 = (n-5)(n+4) = 0
n = 5
Answer: 5, 7