system of quadratic equations

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homeschool girl

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Consider the system of quadratic equations

[MATH] y =3x^2 - 5x,\\ y = 2x^2 - x - c, [/MATH]
where [MATH]c[/MATH] is a real number.


(a) For what value(s) of [MATH]c[/MATH] will the system have exactly one solution [MATH](x,y)?[/MATH]
(b) For what value(s) of [MATH]c[/MATH] will the system have more than one real solution?

(c) For what value(s) of [MATH]c[/MATH] will the system have no real solutions?

tried many things, including the quadratic equation, and setting both equations equal to each other, but none of them worked.
(pls don't ask me to show all the calculations I did, b/c I filled three full sheets of paper and it would take hours to write them down)
 
homeschool girl said:
Consider the system of quadratic equations
[MATH] y =3x^2 - 5x,\\ y = 2x^2 - x - c, [/MATH] where [MATH]c[/MATH] is a real number.
(a) For what value(s) of [MATH]c[/MATH] will the system have exactly one solution [MATH](x,y)?[/MATH](b) For what value(s) of [MATH]c[/MATH] will the system have more than one real solution?
(c) For what value(s) of [MATH]c[/MATH] will the system have no real solutions?
Click to expand...
If we subtract we get \(x^2-4x+c\).
Using the discriminate,\(\delta\), what value of \(c\) makes \(\delta=0,~>0,~<0~?\)
 
Please pick one attempt, and show your work, so we can see how close you came, and whether you just made a small mistake.

I would start as you said by setting the two expressions for y equal (or, equivalently, subtracting one equation from the other), and then using the quadratic formula (or just the discriminant). I suspect that you may have missed the fact that you are looking only for the number of solutions, which is what the discriminant tells you.
 
when I tried using the quadratic equation

[MATH]y=3x^2-5x[/MATH]
[MATH]3x^2-5x-y=0[/MATH]
[MATH]x=\frac{5\pm \sqrt{-5^2-4\cdot3\cdot y}}{2\cdot3}[/MATH]
[MATH]x=\frac{5\pm \sqrt{25-12y}}{6}[/MATH]
[MATH]y=2x^2-x-c[/MATH]
[MATH]2x^2-x-c-y[/MATH]
[MATH]x=\frac{-1\pm \sqrt{1^2-4\cdot2\cdot (-c-y)}}{2\cdot2}[/MATH]
[MATH]x=\frac{-1\pm \sqrt{1+8c+8y}}{4}[/MATH]
[MATH]x=\frac{-1\pm 1 + 2\sqrt{2}c +2\sqrt{2}y}{4}[/MATH]
[MATH]x=\frac{\sqrt{2}c +\sqrt{2}y}{2}[/MATH]
in [MATH]x=\frac{5\pm \sqrt{25-12y}}{6}[/MATH] if [MATH]y>2[/MATH] then it will not be real.

but when I try to put the value of y into [MATH]x=\frac{\sqrt{2}c +\sqrt{2}y}{2}[/MATH] I hit a dead end.
 
[MATH]\delta[/MATH] is the small letter delta in the Greek alphabet. Greek letters are used frequently in mathematics.

The first five letters in that alphabet are alpha, beta, gamma, delta, and epsilon, shown as small letters as

[MATH]\alpha,\ \beta, \ \gamma, \delta, \text { and } \epsilon.[/MATH]
The general method for solving simultaneous equations is substitution: you use one equation to find y in terms of x and then replace y in the other equation with the expression for y in terms of x. In this problem, that has already been done.

[MATH]y = 3x^2 - 5x \text { and } y = 2x^2 - x + c \implies 3x^2 - 5x = 2x^2 - x + c \implies x^2 - 4x + c = 0.[/MATH]
So what does x equal?

When does that result in two distinct real solutions for x? One real solution? No real solutions?

PS Notice the relationship between our word "alphabet" and the names of the first two letters in the Greek version. Notice as well that four of the Greek symbols for small letters look a lot like their equivalents in the Roman alphabet. They represent the same sounds in both alphabets. The gamma does not look like anything in our alphabet, but represents the sound we show as g.
 
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I don't understand how getting c=4 and x=2 correlates with the original problem, and my class hasn't talked about delta yet and I don't know if I should use it in my answer.
 
Don't worry about the symbol used; have you heard of the discriminant, [MATH]b^2 - 4ac[/MATH], and how it tells you how many solutions the quadratic equation has?

You want to choose c so that the equation [MATH]x^2 - 4x + c = 0[/MATH] has exactly one solution; that means [MATH]b^2 - 4ac[/MATH] must be 0.
 
yes, I've learned about discriminant. thank you so very much I was feeling extremely frustrated and completely lost. What you wrote makes sense
 
Dr.Peterson said:
Don't worry about the symbol used; have you heard of the discriminant, [MATH]b^2 - 4ac[/MATH], and how it tells you how many solutions the quadratic equation has?

You want to choose c so that the equation [MATH]x^2 - 4x + c = 0[/MATH] has exactly one solution; that means [MATH]b^2 - 4ac[/MATH] must be 0.
Click to expand...
so for both discriminets to be zero y = [MATH]\frac{25}{12}[/MATH]and c = [MATH]-\frac{53}{24}[/MATH]?
 
No; there's only one discriminant to work with, and you will not be finding a value of y. I don't think you did what we've suggested.

Do you see that if [MATH]y = 3x^2 - 5x[/MATH] and [MATH]y = 2x^2 - x - c[/MATH], then [MATH]3x^2 - 5x = 2x^2 - x - c[/MATH], and therefore [MATH]x^2 - 4x + c = 0[/MATH]? In order for the system to have exactly one solution, the discriminant of this last equation has to be 0; that will give you an equation to solve for c. (The answer is very simple, not an unpleasant fraction.)

Please show the steps of your work, rather than just your conclusion; we need to see where you are going wrong.
 
so its a diffrence of squares and x=2 and c=4?

[MATH]x^2 - 4x + c = 0[/MATH][MATH]x^2 - 4x + 4 = 0[/MATH][MATH](x-2)^2=0[/MATH]
 
OK, the discriminant of [MATH]x^2 - 4x + c[/MATH] is [MATH]16 - 4c[/MATH]; this is zero when [MATH]c = 4[/MATH], as you have found. This also implies that the quadratic is a perfect square, so that there is one solution. You could also use what you have shown (completing the square) to determine c without using the discriminant; perhaps that is what you did to determine that [MATH]c = 4[/MATH].

To answer the question, it isn't necessary to actually find the solution, but you now know that the one solution will be at [MATH]x = 2[/MATH], and you could find y if you want (from the original equations, which will give the same value).

Now, how about parts (b) and (c)?
 
for (b),
[MATH]x^2 - 4x + c = 0[/MATH] would have to not be a perfect square?

and for (c) the discriminant has to be negative?
 
homeschool girl said:
so its a diffrence of squares and x=2 and c=4?

[MATH]x^2 - 4x + c = 0[/MATH][MATH]x^2 - 4x + 4 = 0[/MATH][MATH](x-2)^2=0[/MATH]
Click to expand...
You are almost there. Let's review.

With respect to simultaneous equations, here are the rules.

You cannot get a finite number of numerical solutions if the number of equations is less than the number of variables. (There is an exception for things like Diophantine equations, but they are an advanced topic.)

You cannot get any solutions if the equations are inconsistent.

You may be able to get a finite number of numerical solutions if the number of independent equations is at least equal to the number of variables and the equations are all consistent.

So you have three variables in this problem, but only two equations. Therefore you cannot find numerical solutions for x, y, and c. You can reduce the system to a single equation in two variables by substitution, but you cannot find numerical solutions to that equation either because 1 < 2.

Clear?

What you can do is to find what values of c will give you two real solutions for x, one real solution for x, or no real solution for x.

Why can you do that?

Because the number of real solutions to a quadratic depends on the sign of the discriminant, and there is only one variable in the discriminant.

So you will not know the numerical solution or solutions to x, but you can determine what range of values for c will make the discriminant negative, zero, or positive.

With me so far?

Based on that, you can determine what range of values for c will result in two, one, or no real solutions for x.

You have found the unique value for c, namely c = 4, that makes the discriminant zero, and in that special case, you can now find numeric solutions for x and y because NOW c is fixed rather than variable. But you will not find unique values for c that make the discriminant positive, so c remains a variable, so you will not be able to find numerical values for x and y in those cases. But the problem does not ask you even to try doing that impossibility. Tricky problem. Glad you persisted.

It is a long train of reasoning so let us know if you get lost somewhere.
 
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but I don't get whether I am I trying to get the discriminant or solve [MATH]x^2-4x+c[/MATH]?
i'm so confused now
 
homeschool girl said:
but I don't get whether I am I trying to get the discriminant or solve [MATH]x^2-4x+c[/MATH]?
i'm so confused now
Click to expand...
In general, you cannot solve [MATH]x^2 - 4x + c = 0[/MATH]
in the sense of finding a finite number of numerical values for c and x.

In the special case where there is only one real solution for x, you can find c, x, and y because there is a third equation implied by the fact of just one real solution.

BUT THE QUESTION DOES NOT ASK YOU TO FIND SPECIFIC VALUES FOR x, y, and c, does it? It askes for ranges of values for c.
 
homeschool girl said:
when I tried using the quadratic equation

[MATH]y=3x^2-5x[/MATH]
[MATH]3x^2-5x-y=0[/MATH]
[MATH]x=\frac{5\pm \sqrt{-5^2-4\cdot3\cdot y}}{2\cdot3}[/MATH]
[MATH]x=\frac{5\pm \sqrt{25-12y}}{6}[/MATH]
[MATH]y=2x^2-x-c[/MATH]
[MATH]2x^2-x-c-y[/MATH]
[MATH]x=\frac{-1\pm \sqrt{1^2-4\cdot2\cdot (-c-y)}}{2\cdot2}[/MATH]
[MATH]x=\frac{-1\pm \sqrt{1+8c+8y}}{4}[/MATH]
[MATH]x=\frac{-1\pm 1 + 2\sqrt{2}c +2\sqrt{2}y}{4}[/MATH]
[MATH]x=\frac{\sqrt{2}c +\sqrt{2}y}{2}[/MATH]
in [MATH]x=\frac{5\pm \sqrt{25-12y}}{6}[/MATH] if [MATH]y>2[/MATH] then it will not be real.

but when I try to put the value of y into [MATH]x=\frac{\sqrt{2}c +\sqrt{2}y}{2}[/MATH] I hit a dead end.
Click to expand...
No, this is not correct at all. You are missing a very important point.

Let's go over this slowly.

You have y = 3x^2-5x. If for some value of x, 3x^2-5x = 11 than what would y equal for that same x? The answer is 11. This is because y equals what ever 3x^2-5x equals. That is why y = 3x^2-5x. Again if you some x value, 3x^2-5x = 223 then what would y equal for that same x value. Answer is 223.

Since y is the same as 3x^2-5x, we have 3x^2-5x - y always equaling 0, ie 3x^2-5x-y = 0 for any x value.

What you were trying to do was to find when 3x^2-5x equals 0. You could have said you were trying to find out when y=0. They are the same statement.

Since you want to find out when 3x^2-5x = 0, there is no y in this equation. Just use the quadratic formula on 3x^2-5x=0, where a=3, b=-5 and c=0. Why is c=0? Because you can write 3x^2-5x as 3x^2-5x + 0.

Having said all this I would not proceed in this way. I would set the two equations equal to zero.


Since I really want you to study your mistake about using the quadratic formula for 3x^2-5x-y =0 (please please, please don't move on until you understand why this is wrong), i will help you along with this problem.

I would set 3x^2-5x = 2x^2-x-c. This gives you x^2-4x + c = 0. The solution to this equation will give you all the x-values of the points of intersection(s)

The discriminant (\(\displaystyle \sqrt{b^2-4ac}\)) determines how many points of intersections there are. Not that the c in \(\displaystyle \sqrt{b^2-4ac}\) and the c in x^2-4x + c = 0 happen to be the same value.

As pka suggested, let's denote \(\displaystyle \sqrt{b^2-4ac}\) by \(\displaystyle \delta\)

If \(\displaystyle \delta = 0\), then there is 1 point of intersection.
If \(\displaystyle \delta >0\), then there are 2 points of intersection.
If \(\displaystyle \delta < 0\), then there are no point of intersection.

I did not read all the posts above so I am sure that this has already been stated. My main purpose for this post was to explain why your method of finding the roots/zero for y=3x^2-5x was not correct. Before you found the roots did you thing knowing the roots would have helped? Why?
 
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It would perhaps have been a better problem if they had given you the two equations plus the fact that there was a unique numerical value for x and y. Then you would have searched for the third equation and found it in

[MATH]16 - 4c = 0.[/MATH]
Then they could have asked the follow up question of what values (plural) if x has two real solutions.
 
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