Finding zeros (math homework)

[VEN]

I missed some school and therefore some school material. So my question is, how do you find the zeros of a quadratic relation without using a graphing calculator?

Thanks

[Pat Fairbank]

You use the quadratic formula:

x = (-b +- sqrt(b^2 - 4ac))/(2a)

(where a, b, and c refer to the coefficients of the equation in the form y = ax^2 + bx + c)

[VEN]

So say my relation is A = w(20-w), I'd solve that by making w=0 and 20-w = 0? So all this is, is substitution? Guess and check?

[lukevanoort]

You can also factor it out into two quantities. Eg. x^2+3x+2=0 factors out into (x+1)*(x+2)=0 since one of those quantities must equal zero and it could be either one you set both equal to zero, so it would be x+1=0 and x+2=0. These solve out to x=-1 ans x=-2. (-1)^2+3(-1)+2 goes to 1+-3+2 which proves the x=-1 answer. Then we do the other one, (-2)^2+3(-2)+2 goes to 4+-6+2 which equals zero too. So now we are 100% sure that it is x=-2 and x=-1.

[lukevanoort]

Originally Posted by VEN

So say my relation is A = w(20-w), I'd solve that by making w=0 and 20-w = 0? So all this is, is substitution? Guess and check?

Well you distribute w(20-w) to 20w-w^2, then subtract A, so you get
-w^2+20w-A=0. Then you use the quadratic formula so it goes in as
(-20 (+/-) sqrt(400-4(-1)(-A)))/-2 which simplifies to (-20 (+/-) sqrt(400-4A))/-2 you can factor a 4 out of (400-4A) so it becomes 4(100-A). The sqrt of 4 is two, so you get (-20 (+/-) 2 sqrt(100-A))/-2. You can divide the -20 and the 2 by the -2 so your final answer is 10 (+/-) sqrt(100-A).

PS Can someone check me on that? I'm a bit sleepy.

[VEN]

here's my last question, I hope:

A pizza company's research shows that a 25 cent increase in the price of a pizza results in 50 fewer pizzas being sold. The usual price of $15 for a pizza results in sales of 1000 pizzas. The questions are a) Write the algerbraic expression that models the revenue for this situation. and b) what is the optimum Max value?

so far I wrote down on my sheet y=(15+.25x)(1000-50x) This models the revenue so I guess I answered the first question

I know I can solve the second question by using a table and writing down the points of the parabola until I reach the max value, but how do I do it using the algebraic method?

[Pat Fairbank]

Originally Posted by VEN

So say my relation is A = w(20-w), I'd solve that by making w=0 and 20-w = 0? So all this is, is substitution? Guess and check?

Let me explain the whole derivation of the quadratic formula.

You have a quadratic equation of the form y = ax^2 + bx + c. You want to find the roots (zeros) of the equation, meaning the values of x at which y is 0. So the equation takes the form ax^2 + bx + c = 0. (Or if the equation already has this form, you can start from here.)

Now, all we have to do is solve for x.

First, you factor the a out of ax^2 + bx, and put the c on the other side:

a[x^2 + (b/a)x] = -c

Next, add b^2/(4a^2) inside the square brackets to complete the square (gotten by dividing b/a by 2 and squaring). Multiply this value by a and add it to the other side to equalize things:

a[x^2 + (b/a)x + b^2/(4a^2)] = b^2/(4a) - c

The contents of the square brackets can now be reduced to [x + b/(2a)]^2, and at the same time we'll make the right side into one fraction:

a[x + b/(2a)]^2 = (b^2 - 4ac)/(4a)

Divide both sides by a:

[x + b/(2a)]^2 = (b^2 - 4ac)/(4a^2)

Take the square root of both sides:

x + b/(2a) = +-sqrt(b^2 - 4ac)/(2a)

Isolate x:

x = -b +-sqrt(b^2 - 4ac)/(2a)

And now you have a formula that can solve for x whenever you have an equation that looks like 0 = ax^2 + bx + c.

[VEN]

grr.....why does it have to be so confusing.....at least to me

Thanks though. I'll try and understand

[Pat Fairbank]

Originally Posted by VEN

here's my last question, I hope:

A pizza company's research shows that a 25 cent increase in the price of a pizza results in 50 fewer pizzas being sold. The usual price of $15 for a pizza results in sales of 1000 pizzas. The questions are a) Write the algerbraic expression that models the revenue for this situation. and b) what is the optimum Max value?

so far I wrote down on my sheet y=(15+.25x)(1000-50x) This models the revenue so I guess I answered the first question

I know I can solve the second question by using a table and writing down the points of the parabola until I reach the max value, but how do I do it using the algebraic method?

The maximum value of a parabola is found right between the two zeros, so the easy way to derive a formula for the x-value of the maximum, if you know the quadratic formula, is to find the average of the two zeros:

x = ([-b + sqrt(b^2 - 4ac)/(2a)] + [-b - sqrt(b^2 - 4ac)/(2a)]) / 2

This reduces to x = -b/(2a).

So to answer the pizza question you need to multiply out your equation into the form y = ax^2 + bx + c, and apply that formula. That should give you the number of price increases.

[VEN]

Ok, I think I got the max value question. I do:

15+.25x=0 1000-50x=0
15=-.25x 1000=50x
x=15/(-.25) x=1000/50
x=-60 x=20

now I find the mid point:

(-60+20)/2=-20 which gives me my x value of the vertex. grrrrr...................

[Alex Cormier]

Maybe you should try Division by zero.

[VEN]

I'm so mad at the moment that I'm debating on whether I should move to Ottawa for the summer and ask my dad's sister to teach me math...and I think I'll go to this after school math thing for the next week because if I don't get this AND I'm going to Atlanta, I'll be a total numbskull when I get back

THANKS FOR ALL THE HELP GUYS. I managed to do something with my homework