# Quadratic Equation - Nature of roots

• Jan 17th 2010, 06:47 AM
Punch
Quadratic Equation - Nature of roots
(a)Find the value of m for which the line $y=mx-3$ is a tangent to the curve $y=x+\frac{1}{x}$ and find the x-coordinate of the point at which this tangent touches the curve.

(b) Find the value of c and of d for which $(x:-5 is the solution set of $x^2+cx

For (a), how do I actually solve the quadratic equation when the equation have a $x^{-1}$?

No idea how to solve (b)... The whole question is just like alien to me....
• Jan 17th 2010, 07:01 AM
Bacterius
This has nothing to do with quadratic equations. This is not algebra but calculus.
You have to take the derivative $y'$ of $y$ :

$y = x + \frac{1}{x} = x + x^{-1}$

Thus, following first principles, $y' = 1 - x^{-2}$

Have you done derivatives ? Can you see how to do it now ?
• Jan 17th 2010, 07:02 AM
Quote:

Originally Posted by Punch
(a)Find the value of m for which the line $y=mx-3$ is a tangent to the curve $y=x+\frac{1}{x}$ and find the x-coordinate of the point at which this tangent touches the curve.

(b) Find the value of c and of d for which $(x:-5 is the solution set of $x^2cx

For (a), how do I actually solve the quadratic equation when the equation have a $x^{-1}$?

No idea how to solve (b)... The whole question is just like alien to me....

hi punch ,

for (1) , you multiply the whole thing by x so you don end up with $x^{-1}$

$mx-3=x+\frac{1}{x}$

$mx^2-3x=x^2+1$

$(m-1)x^2-3x-1=0$ **

If a line is a tangent to the curve , it touches the curve at one point .

so $b^2-4ac=0$

$(-3)^2-4(m-1)(-1)=0$

calculate the value of m , then put this value of m into ** , evaluate the x-coordinate then .

for (2) , you are given the range , so

$(x+5)(x-3)<0\Rightarrow x^2+2x-15<0$

compared this with , $x^2+cx-d<0$ ?? (i think you have a typo here)

we see that c=2 , d=15 assuming that i guess it correctly .
• Jan 17th 2010, 07:05 AM
Bacterius
Forgive my last post, I guess there must be a way to do it with algebra ... the hard way.
Forgive also what I said in this post, lol, I must be tired.
As mathaddict said, try to multiply the equations to get rid of the fractions and revert back to the good old quadratic equation. It makes it a lot easier.
• Jan 17th 2010, 07:16 AM
Punch
Quote:

hi punch ,

for (1) , you multiply the whole thing by x so you don end up with $x^{-1}$

$mx-3=x+\frac{1}{x}$

$mx^2-3x=x^2+1$

$(m-1)x^2-3x-1=0$ **

If a line is a tangent to the curve , it touches the curve at one point .

so $b^2-4ac=0$

$(-3)^2-4(m-1)(-1)=0$

calculate the value of m , then put this value of m into ** , evaluate the x-coordinate then .

for (2) , you are given the range , so

$(x+5)(x-3)<0\Rightarrow x^2+2x-15<0$

compared this with , $x^2+cx-d<0$ ?? (i think you have a typo here)

we see that c=2 , d=15 assuming that i guess it correctly .

yes sorry for my typo. Well, thanks and I understood part a fully. About part b, it is about turning the x values into equation and then expanding and comparing to find c and d... it just seems so much like any other quadratic equations :) thanks
• Jan 17th 2010, 06:19 PM
HallsofIvy
Quote:

Originally Posted by Punch
(a)Find the value of m for which the line $y=mx-3$ is a tangent to the curve $y=x+\frac{1}{x}$ and find the x-coordinate of the point at which this tangent touches the curve.

(b) Find the value of c and of d for which $(x:-5 is the solution set of $x^2+cx

For (a), how do I actually solve the quadratic equation when the equation have a $x^{-1}$?

What quadratic equation are you talking about? There is no quadratic equation here. A line crosses a graph if the equation of the line and the equation of the graph have a common factor- that is, if $mx- 3= x+ \frac{1}{x}$. The line is tangent to the curve if that is a double root. To get a quadratic equation, multiply both sides of the equation by x.

For (b), start by finding solutions to the equation $x^2+ cx= d$. Those two points must separate "<" from ">". Try a single point, say x= 0, to determine which is which.
• Jan 18th 2010, 12:50 AM
Punch
Sorry. the "quadratc equation i was talking about was an equation with x to the power of -1, the question have been solved.