Find Value of p: Solve Quadratic Equation

[utkarshakash]

Homework Statement

If (x^2 -x+p)(11y^2 - 4y + 2) = \frac{9}{2} have exactly one ordered pair of (x,y) then the value of p is .....

The Attempt at a Solution

Do I have to find the ordered pair in order to solve this question? Also the quadratic in y does not seem to have real roots. Does this mean that the quadratic in x should also have complex roots?

 

[SammyS]

Homework Statement

If (x^2 -x+p)(11y^2 - 4y + 2) = \frac{9}{2} have exactly one ordered pair of (x,y) then the value of p is .....

The Attempt at a Solution

Do I have to find the ordered pair in order to solve this question? Also the quadratic in y does not seem to have real roots. Does this mean that the quadratic in x should also have complex roots?

Your observation that \ 11y^2 - 4y + 2\ does not have real roots is very helpful.

Where is the vertex of this quadratic located? Does that give a minimum or does it give a maximum ?

 

[utkarshakash]

Your observation that \ 11y^2 - 4y + 2\ does not have real roots is very helpful.

Where is the vertex of this quadratic located? Does that give a minimum or does it give a maximum ?

At y=2/11, the minimum occurs.

 

[ehild]

You have the product of two functions f(x) and g(y). Assuming that Edit: f(x)= [STRIKE]p[/STRIKE] r and g(y)=q, rg=9/2. It is required that there is only one pair of x and y that satisfies the equation. Both f(x) and g(y) are quadratic functions. Usually there are two x values so as f(x)=r, and the same with g(y)=q, except ?

ehild

 

[utkarshakash]

You have the product of two functions f(x) and g(y). Assuming that f(x)= p and g(y)=q, pg=9/2. It is required that there is only one pair of x and y that satisfies the equation. Both f(x) and g(y) are quadratic functions. Usually there are two x values so as f(x)=q, and the same with g(y), except ?

ehild

I think taking f(x)=p is a bit ambiguous as the quadratic already contains p. I also have a doubt regarding your last statement. Are you trying to say this:

" Usually there are two x values so as f(x)=p"

rather than

" Usually there are two x values so as f(x)=q "

 

[ehild]

I think taking f(x)=p is a bit ambiguous as the quadratic already contains p. I also have a doubt regarding your last statement. Are you trying to say this:

" Usually there are two x values so as f(x)=p"

rather than

" Usually there are two x values so as f(x)=q "

You are right, I made a big confusion by using the notation p again. So assume that f(x) =x²-x+p=r, g(y)=11y²-4y+2=q, and rq=9/2. f(x)=r can be valid only for a single x, and the same holds for g(y)=q.

ehild

 

[haruspex]

At y=2/11, the minimum occurs.

So what is the minimum value of 11y²-4y+2? Going back to the original equation, what does that tell you about the other factor?

 

[utkarshakash]

So what is the minimum value of 11y²-4y+2? Going back to the original equation, what does that tell you about the other factor?

The minimum value is 18/11. Also the minimum of the other factor occurs at x=1/2. Are you trying to say that the ordered pair in this case is x=1/2, y=2/11? OK, if for a moment I assume this is indeed true, fortunately this gives me the answer. But I'm still not satisfied as why does the ordered pair has to be point of minima in this case!

 

[haruspex]

The minimum value is 18/11. Also the minimum of the other factor occurs at x=1/2. Are you trying to say that the ordered pair in this case is x=1/2, y=2/11? OK, if for a moment I assume this is indeed true, fortunately this gives me the answer. But I'm still not satisfied as why does the ordered pair has to be point of minima in this case!

I didn't ask what value of x minimises the other term. I asked what the minimum value of the 'y' factor is, and what combining that fact with the original equation tells you about the 'x' factor.

However, it is possible to do it your way too. Suppose p is not the value you computed. Substituting the minimising (x, y) pair will now not make the equation balance. Suppose the RHS is the smaller. Since we chose x and y to minimise the LHS we will not be able to find another (x, y) that fits the equation, so that cannot be.
So the LHS must be the smaller. Now, could we make the equation balance by changing just x? Would the choice of the new x be unique? Could we make the equation balance by changing just y?

 

[ehild]

The problem can be written as f(x)g(y)=4.5. At a given (x1,y1) both f and g have some value. Say, f(x1)=a and g(y1)=b. ab=4.5. Both f(x) and g(y) are quadratic functions. If you solve the quadratic equation g(y)=11y²-4y+2=b you get two roots

y_{1,2}=\frac{2\pm \sqrt{4-11(2-b)}}{11}, except when the discriminant is zero.

The same for f(x). If you solve the equation x²-x+p=a, you get two roots again, except when the discriminant is zero.
x_{1,2}=\frac{1\pm \sqrt{1-4(p-a)}}{2}.

That means if (x1,y1) is a solution, (x1,y2), (x2,y1), (x2,y2 ) are also solutions. But the solution must be unique, all these solutions must be the same, that is x1=x2, y1=y2. The dicriminants must be zero, and that holds when both x1 is at the peak of f(x) and y1 is at the peak of g(y).

ehild

 

[Gzyousikai]

(x^2-x+p)(11y^2-4y+2)=\left[\left(x-\frac{1}{2}\right)^2+p-\frac{1}{4}\right]\left[11\left(y-\frac{2}{11}\right)^2+\frac{18}{11}\right]\ge\left(p-\frac{1}{4}\right)\left(\frac{18}{11}\right)=\frac{9}{4}. solve for p and get p=3

 

[ehild]

Right!

ehild

 

[AlephZero]

(x^2-x+p)(11y^2-4y+2)=\left[\left(x-\frac{1}{2}\right)^2+p-\frac{1}{4}\right]\left[11\left(y-\frac{2}{11}\right)^2+\frac{18}{11}\right]\ge\left(p-\frac{1}{4}\right)\left(\frac{18}{11}\right)=\frac{9}{4}. solve for p and get p=3

The inequality is only true if p - 1/4 >= 0.

But it's a neat trick to get the solution, given that we know a solution exists (because if there was no solution, the question would be nonsense).

 

[haruspex]

(x^2-x+p)(11y^2-4y+2)=\left[\left(x-\frac{1}{2}\right)^2+p-\frac{1}{4}\right]\left[11\left(y-\frac{2}{11}\right)^2+\frac{18}{11}\right]\ge\left(p-\frac{1}{4}\right)\left(\frac{18}{11}\right)=\frac{9}{4}. solve for p and get p=3

Isn't that what utkarshakash did, in effect, in post #8?