**Find all fixed points for the following functions by setting the function equal to x and solving the resulting equation for x**

**Application of quadratic equation and quadratic functions**

(1) f(x)=x^2-9x+25

(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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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

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