Quadratic & Equation of a Circle

[Physiona]

Homework Statement

Two equations are defined as follows:

What is the value of

?

Homework Equations

Quadratic Format: $ax^2+bx+c=0$
$y^-x$ = $1/y^x$

The Attempt at a Solution

I'm not sure In how to attempt this style of question as I know quadratic and equation of a circle fully. (This question has the quadratic in a different format)[/B]
I've attempted to rearrange the second equation into making y the subject:
$3y= 2x-3$
$y= 2x-3/3$
I've substituted it in the second equation in place of $y$.
$x^2+((2x/3)/3)^2 = 45$
Led me to this quadratic.. $13x^2-12x-396=0$.
This way doesn't look right to me, though my teacher applied me with this methodology. Any guidance as if it's correct or not and how I should progress into getting the right answer?

 

[Ray Vickson]

Homework Statement

Two equations are defined as follows:

View attachment 223079

View attachment 223080

What is the value of View attachment 223081?

Homework Equations

Quadratic Format: $ax^2+bx+c=0$
$y^-x$ = $1/y^x$

The Attempt at a Solution

I'm not sure In how to attempt this style of question as I know quadratic and equation of a circle fully. (This question has the quadratic in a different format)[/B]
I've attempted to rearrange the second equation into making y the subject:
$3y= 2x-3$
$y= 2x-3/3$
I've substituted it in the second equation in place of $y$.
$x^2+((2x/3)/3)^2 = 45$
Led me to this quadratic.. $13x^2-12x-396=0$.
This way doesn't look right to me, though my teacher applied me with this methodology. Any guidance as if it's correct or not and how I should progress into getting the right answer?

How do you go from $y = 2x - 3/3 \Rightarrow y = 2x - 1$ from the linear equation to $y^2 = ((2x/3/3)^2$ in the circle equation?

 

[Physiona]

How do you go from $y = 2x - 3/3 \Rightarrow y = 2x - 1$ from the linear equation to $y^2 = ((2x/3/3)^2$ in the circle equation?

I don't get where the $y^2 = ((2x/3/3)^2$ has come from the circle equation; because isn't this meant to be in place of $y$?
In order to go from $y = 2x - 3/3 \Rightarrow y = 2x - 1$ you rearrange the first equation in the format to becoming $3y = 2x-3$ which becomes $y=2x/3/3$... Right...

 

[fresh_42]

You should have solved your quadratic equation. I further assume, that we are only interested in real values?

 

[Mark44]

I've substituted it in the second equation in place
of $y$.
$x^2+((2x/3)/3)^2 = 45$
Led me to this quadratic.. $13x^2-12x-396=0$.

The quadratic looks fine, but the first equation should be $x^2 + ( (2/3)x - 1)^2 = 45$.

 

[Physiona]

The quadratic looks fine, but the first equation should be $x^2 + ( (2/3)x - 1)^2 = 45$.

Okay, but why is the first equation $x^2 + ( (2/3)x - 1)^2 = 45$?

 

[Physiona]

You should have solved your quadratic equation. I further assume, that we are only interested in real values?

Is my quadratic and attempt of the question right...?

 

[fresh_42]

Is my quadratic and attempt of the question right...?

You should have tried. One solution of the two will lead to the answer. It's a line intersecting a circle, so only two possibilities.

 

[Physiona]

Yes I got two solutions,

$x=6$ and $x=−5.07692$
I require y though. I'm presuming I substitute one value in the equation to find y as I need that. Which equation to choose and which value is tricky.

You should have tried. One solution of the two will lead to the answer. It's a line intersecting a circle, so only two possibilities.

 

[fresh_42]

Which equation to choose...

Guess which: the linear or the quadratic. One of them sounds easier than the other.

... and which value is tricky.

That's why I asked for a real or (many) complex solutions. You should calculate with $-\frac{66}{13}$ instead of the decimal number. Then you will at least get both solutions formally. But I guess only one is requested.

 

[Physiona]

Guess which: the linear or the quadratic. One of them sounds easier than the other.

That's why I asked for a real or (many) complex solutions. You should calculate with $-\frac{66}{13}$ instead of the decimal number. Then you will at least get both solutions formally. But I guess only one is requested.

I presume the linear is easier.
The value out of the ones I've calculated, clearly $x=6$ is easier to use. And I obtain y which I then use on to figure out $y^-x$, correct?

 

[fresh_42]

Yes. I would calculate both. So in principle you'll know all other solutions, too, or can at least write them down, depending on your knowledge of complex roots.

 

[epenguin]

Probably faster than all of this 45 is the sum of two squares. Can you think of two simple whole numbers whose squares add up to 45?

(Are you sure the question was y^-x and not y-x ?)

 

[Ray Vickson]

I don't get where the $y^2 = ((2x/3/3)^2$ has come from the circle equation; because isn't this meant to be in place of $y$?
In order to go from $y = 2x - 3/3 \Rightarrow y = 2x - 1$ you rearrange the first equation in the format to becoming $3y = 2x-3$ which becomes $y=2x/3/3$... Right...

I don't get it either, but I just copied it exactly from what you wrote.