How Do You Solve These Quadratic Equations in Real-Life Scenarios?

In summary, the conversation involves two problems - one involving a quadratic equation to find the number of shares bought and the other involving a system of linear equations to find the lengths of a rectangular field's sides. The conversation also includes a question about finding the value of a variable in a given equation.
  • #1
TheShapeOfTime
I was given a bunch of quadratic homework and I'm stuck on a few problems:

"Brendan buys a black of shares for $1895. When the share price goes up by $4/share, he sells all but 15 of them for $1740. How many shares did he buy?"

I got the equation [tex]\frac{1895}{x} - \frac{1740}{x + 4} = 15[/tex] but I worked it out and I don't get the answer (~73).

--

"A rectangular field has a perimeter 500 m and an area 14 400 m^2. Find the lengths of its sides."

[tex]l * w = 14 400[/tex]
[tex]2l + 2w = 500[/tex]
Not sure what to do from here...

--

Thanks in advance for any help
 
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  • #2
How on Earth did you get your first equation??
It is, as far as I can see, totally wrong.

Let "P" be the initial price for one share, and "x" the number of shares originally bought.
Then, the information given can be written as:
P*x=1895
(P+4)*(x-15)=1740

We therefore have [tex]P=\frac{1895}{x}[/tex]
This yields a quadratic in x you can solve.
 
  • #3
And the second problem you have set up right and is now simply solving a system of linear equations. Solve one of the eqns for one variable ... substitute ... sound familiar?
 
  • #4
t!m said:
And the second problem you have set up right and is now simply solving a system of linear equations. Solve one of the eqns for one variable ... substitute ... sound familiar?

Yeah, I had worked it out like that earlier and didn't get the right answer, but I realize now it was just a dumb mistake.

There is one last part of a question I can't get:

"Given [itex]kx^2+(k+3)x+(3-4k) = 0[/itex]. Find the value of k if
c) one of the roots is 2"

Thanks for your help, both of you.
 
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  • #5
Notice that in the first equation you gave:[tex]\frac{1895}{x} - \frac{1740}{x + 4} = 15[/tex], The numerator of each fraction on the left is an amount of money: dollars, while the denominator is the number of shares bought: each fraction is "dollars per share".

Actually, that's not quite correct because I assume you got the "4" from "$4/share" and you can't add "shares" to "dollars per share". You CAN subtract "shares" from "shares", of course, so I feel sure that what you intended was the second denominator to be "x- 15": the number of shares less the number NOT sold: the number of shares sold. Of course, then, the right hand side of the equation must also be in "dollars per share": $4/share would work nicely. Looks to me like the equation you intended to write was [tex]\frac{1895}{x} - \frac{1740}{x -15} = 4[/tex].
Solve that for x.
 
  • #6
HallsofIvy said:
Notice that in the first equation you gave:[tex]\frac{1895}{x} - \frac{1740}{x + 4} = 15[/tex], The numerator of each fraction on the left is an amount of money: dollars, while the denominator is the number of shares bought: each fraction is "dollars per share".

Actually, that's not quite correct because I assume you got the "4" from "$4/share" and you can't add "shares" to "dollars per share". You CAN subtract "shares" from "shares", of course, so I feel sure that what you intended was the second denominator to be "x- 15": the number of shares less the number NOT sold: the number of shares sold. Of course, then, the right hand side of the equation must also be in "dollars per share": $4/share would work nicely. Looks to me like the equation you intended to write was [tex]\frac{1895}{x} - \frac{1740}{x -15} = 4[/tex].
Solve that for x.

You're right, that was what I was going for. I had a typo in it earlier (x + 1 instead of x + 4), which was what made arildno react the way he did :p
 
  • #7
TheShapeOfTime said:
You're right, that was what I was going for. I had a typo in it earlier (x + 1 instead of x + 4), which was what made arildno react the way he did :p
Yeah, I guess I overreacted a bit. Sorry..
 

FAQ: How Do You Solve These Quadratic Equations in Real-Life Scenarios?

What is a quadratic equation?

A quadratic equation is a polynomial equation of the second degree, meaning it contains a variable raised to the power of two. It can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.

How do you solve a quadratic equation?

There are various methods for solving quadratic equations, including factoring, completing the square, and using the quadratic formula. The method to use depends on the specific equation and the tools available to solve it.

What is the discriminant of a quadratic equation?

The discriminant is a value derived from the coefficients of a quadratic equation, given by b^2 - 4ac. It is used to determine the nature of the solutions of the equation. If the discriminant is positive, the equation has two distinct real solutions; if it is zero, the equation has one real solution; and if it is negative, the equation has no real solutions.

How do you graph a quadratic equation?

To graph a quadratic equation, you can plot points by substituting different values of x into the equation and solving for y. Another method is to use the vertex form of the equation, which is y = a(x - h)^2 + k, where (h,k) is the vertex of the parabola. The vertex can also be found by using the formula -b/2a for the x-coordinate and then plugging it into the equation to find the y-coordinate.

What are the real-world applications of quadratic equations?

Quadratic equations are used in various fields, including physics, engineering, and economics. They can be used to model the motion of objects under the influence of gravity, calculate the optimal production level in a business, and determine the trajectory of projectiles, among other things.

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