I know I have to use the equation:

-b (plus or negative)(sqrt)b²-4ac
x= ---------------------------------
2a

The question I have is x²-2x-5=10. I have ot use the completed square form to solve the equation but im confused.

Also another question I have is to investigate by substituting values for x satisfying -5(Larger than or equal to) x (Lessthan or equal to) 5, find the largest possible value of:

6
--------
x²-2x-5

And then justify why my chosen value of x maximises the equation.

Sorry for the long questions and stuff but if oyu wouldnt mind explaining how you did them that would help me pestering you next time

2. Originally Posted by Yppolitia
The question I have is x²-2x-5=10. I have ot use the completed square form to solve the equation but im confused.
x^2 - 2x - 5 = 0

x^2 - 2x = 5

Now, we want to add a number to both sides such that the LHS will be a perfect square. As it happens x^2 + 2a + a^2 = (x + a)^2. Comparing the coefficient of the linear term of this form with the linear term in the equation above we can see that
2a = -2
a = -1

Thus a^2 = (-1)^2 = 1. So add 1 to both sides:
x^2 - 2x + 1 = 5 + 1
(x - 1)^2 = 6
x - 1 = (+/-)sqrt(6)
x = 1 (+/-) sqrt(6)

We can check this using the quadratic formula:
x^2 - 2x - 5 = 0
Gives
x = {-(-2) (+/-) sqrt([-2]^2 - 4*1*(-5))}/(2*1)

x = {2 (+/-) sqrt(24)}/2 = {2 (+/-) sqrt(4*6)}/2 = {2 (+/-) 2*sqrt(6)}/2

x = 1 (+/-) sqrt(6) (Check!)

-Dan

3. Originally Posted by Yppolitia
Also another question I have is to investigate by substituting values for x satisfying -5(Larger than or equal to) x (Lessthan or equal to) 5, find the largest possible value of:

6
--------
x²-2x-5

And then justify why my chosen value of x maximises the equation.
I'm not sure I understand the question. According to the first problem the denominator of this function goes to zero for x = 1 (+/-) sqrt(6) indicating we have vertical asymptotes at these x values. So the "largest possible value" of the function is positive infinity, which really means there is no maximum.

-Dan

4. Originally Posted by topsquark
x^2 - 2x - 5 = 0

x^2 - 2x = 5

Now, we want to add a number to both sides such that the LHS will be a perfect square. As it happens x^2 + 2a + a^2 = (x + a)^2. Comparing the coefficient of the linear term of this form with the linear term in the equation above we can see that
2a = -2
a = -1

Thus a^2 = (-1)^2 = 1. So add 1 to both sides:
x^2 - 2x + 1 = 5 + 1
(x - 1)^2 = 6
x - 1 = (+/-)sqrt(6)
x = 1 (+/-) sqrt(6)

We can check this using the quadratic formula:
x^2 - 2x - 5 = 0
Gives
x = {-(-2) (+/-) sqrt([-2]^2 - 4*1*(-5))}/(2*1)

x = {2 (+/-) sqrt(24)}/2 = {2 (+/-) sqrt(4*6)}/2 = {2 (+/-) 2*sqrt(6)}/2

x = 1 (+/-) sqrt(6) (Check!)

-Dan

I dont get how you've gone from x&#178;-2x-5=10 to x&#178;-2x-5=0

5. Originally Posted by Yppolitia
I dont get how you've gone from x&#178;-2x-5=10 to x&#178;-2x-5=0

x&#178; - 2x - 5 = 10
x&#178; - 2x - 15 = 0

Factorise

(x - 5)(x + 3) = 0

Solutions

x = 5 and x = -3

EDIT
Oh complete the square

(x - 1)&#178; - 1 - 15 = 0
(x - 1)&#178; - 16 = 0
(x - 1)&#178; = 16
x - 1 = + 4 or x - 1 = - 4
x = 5 or x = -3

6. Originally Posted by Yppolitia
I know I have to use the equation:

-b (plus or negative)(sqrt)b²-4ac
x= ---------------------------------
2a

The question I have is x²-2x-5=10. I have ot use the completed square form to solve the equation but im confused.
Hi,

x²-2x = 15

complete the square on the LHS by adding 1. What you do on the LHS you have to do on the RHS too!

x²+2x+1 = 15+1

(x-1)² = 4²

Calculate the square-root on both sides:

x - 1 = 4 or x - 1 = -4

So x = 1 + 4 = 5 or x = 1 - 4 = -3

Originally Posted by Yppolitia
Also another question I have is to investigate by substituting values for x satisfying -5(Larger than or equal to) x (Lessthan or equal to) 5, find the largest possible value of:
6
--------
x²-2x-5

And then justify why my chosen value of x maximises the equation....
As topsquark has already shown this term is undefined for x = 1 +/- sqrt(6). I don't know if you know to calculate a maximum or a minimum by using derivatives. If not, you can use another method:
A fraction which numerator is a constant will be extremly great if the denominator is extremely small. The denominator is a quadratic term. The smallest value correspond to the vertex of a parabola.
x²-2x-5 = x²-2x+1-1-5 = (x-1)³ - 6.

So at x = 1 the smallest value of the quadratic term is -6.

So your fraction has a maximum at x = 1 and the maximum has the value 6/(-6) = -1
I've attached a graph to demonstrate what I've calculated.

7. Originally Posted by topsquark
I'm not sure I understand the question. According to the first problem the denominator of this function goes to zero for x = 1 (+/-) sqrt(6) indicating we have vertical asymptotes at these x values. So the "largest possible value" of the function is positive infinity, which really means there is no maximum.

-Dan
I suspect a mistype in the question as if the minimum point of the function in the denominator was above the x axis then this point would correspond to the maximum of the whole function

8. Originally Posted by Yppolitia
I dont get how you've gone from x²-2x-5=10 to x²-2x-5=0
Sorry. Yes, I misread the question. But if you can follow the method that, at least, is correct. Earboth has it done with the correction.

-Dan